By squaring 25,485 different numbers I have concluded that the square root of 649,485,225 is 25,485
@pietergeerkens6324
4 ай бұрын
From a quick look at 649,485,225, I deduced that its square root is in either [25085, 25985] or [25015, 25915] because: 1) it's general magnitude is 25,000 and something; 2) it's one digit is 5 to give 25 as its final "centit"; and 3) It's ten's digit must be an 8 or a 1, as the only pairs of consecutive natural numbers with a product ending in 2 is 8 * 9 and 1 * 2 (this is the grade school trick that (10n + 5)^2 = 100 * n(n+1) + 25); Then a little binary search gives 25485 after at most 5 trial products. Further, both (b) and (e) are ruled out as divisible by 3 but not 9, having digit sums of respectively 24 and 39, using the old divisibility (and bookkeeping) trick "casting out 9's".
@ZDTF
4 ай бұрын
@@pietergeerkens6324I don't understand that Cuz I'm 12 years old but it looks cool😮
@pietergeerkens6324
4 ай бұрын
@@ZDTF If you have a specific question I will have a go at explanation.
@ZDTF
4 ай бұрын
@@pietergeerkens6324 what is a centit
@The_NSeven
4 ай бұрын
Damn, without thinking too much about it i thought of ~25000 (+5). Wasn't too far off!
@extensivity29
5 ай бұрын
I guessed C without even thinking about it, put me in Oxford
@cookiemonster2298
4 ай бұрын
sanme
@czerwonyvenom80
4 ай бұрын
same
@paulgreen9059
4 ай бұрын
Coming to a conclusion without actually thinking about it sounds more like Harvard. 😀
@chiehlilee9224
4 ай бұрын
Ditto, if you’ve worked with numbers long and often enough, your instinct will give you some ideas.
@betterbee980
4 ай бұрын
I also Guessed C without even looking 😂 so I must be in Oxford. Reason First is not at all Second also no cause 3 is at last Last is also not possible cause 3 zeroes Fourth is not possible cause 35 is at last which is always 25 So only c is left
@Pfisiar22
4 ай бұрын
just check the last 2 digits. a perfect square will never end in 33, , 99, or 35. Perfect squares will only ever end in an even number of 00s. So the only remaining answer is C.
@TOBYSHERIDANWHITEPOWER
4 ай бұрын
Yeah I thought this was quite an obvious way of doing it too?
@mrsig8607
2 ай бұрын
thats what i did, the trickier part is proving that a perfect square will never end in these ways
@miraj2264
5 ай бұрын
For options B and E, their digit sum is 24 and 39 respectively. So both of these numbers are divisible by 3 but not 9 = 3^2. Perfect squares break into a product of primes to even powers so neither B or E can be perfect squares since they only break into (3^1)*(other primes to powers). You can actually use that same rule for option A if you forgot about squares being a certain distance away from each other. 99,999,999 = 10^8 - 1 = (10^4)^2 - 1^2 = (10^4+1)(10^4-1) = 10,001*9,999. 10,001 isn't divisible by 11 but 9,999 is. But how many times? Well 9,999/11 = 909. However, 909 isn't divisible by 11. In other words, this number divides 11 but not 11^2. So it can't be a perfect square either.
@highviewbarbell
5 ай бұрын
Is this like elementary number theory or combinatorics or something? I wanna learn some before I take Discrete this fall
@miraj2264
5 ай бұрын
@@highviewbarbell The former (specifically modular arithmetic). Combinatorics is more along the lines of counting things. For example, you draw 5 cards from a deck of 52 cards. How many ways are there to get a flush?
@LarifariRambazamba
4 ай бұрын
@@highviewbarbell Recursively summing the digits of a base-ten number (aka the "digital root") gives you the residue modulo 9 (except you will get a sum between 1 and 9 instead of between 0 and 8 unless the original number is itself trivially 0). That is because "dₙ…d₁d₀" mod (10−1) = ∑dᵢ10ⁱ mod (10−1) ≡ ∑dᵢ mod (10−1), because 10≡1 mod (10−1) so the 10ⁱ just cancels out. So option E ≡ 3 mod 9, so it is divisible by 3 but not by 9. When recursively summing the digits of 987654000 you can just cross out the initial 9 and the final 54 (5+4 adds up to 9), so 8+7+6=21 and 2+1=3.
@thesquarduck8397
4 ай бұрын
Also a number is divisible by 25 if first 2 digits are divisible by 25 (cuz 100 is already divisible by 25 obviously). So for option D, it is divisible by 5 (ends on 5), but not divisible by 25 == 5^2 (ends on 35). If I knew that from the beginning I could solve this problem in a finger snap, fire idea.
@nicktomato7
4 ай бұрын
simpler for ruling out A is that 99,999,999 is 10^8 - 1, and 10^8 is itself a perfect square: (10^4)^2 even the smallest perfect squares (after 1) are further than 1 away from their neighboring squares, so ‘a perfect square minus 1’ can’t be a perfect square
@nephel6158
4 ай бұрын
I will try doing C by hand. First thing is that I want to get a rough estimate of what the answer should be. I observe that there are 9 digits, so the solution should be a relatively low 5 digit number. Then I tried squaring some easy numbers and adding 0s at the end to try and obtain a range for the solution. So, I observed that 20k^2 is 400M, and 30k^2 = 900M, and C lies inside that range. Then, I try 25k and since 25^2 is 625, 25k^2 is 625M. That's remarkably close to C. So I tried 26^2, which is 676, so 26k^2 = 676M. Next, the last digit of C is 5. so the solution *has* to be a multiple of 5. So, the solution is 25ab5, where a and b are some integers. At this point, I realised I can binary search for a and b and get an answer through about a bit under 10 attempts of manual squaring of 5 digit numbers, but I *really* don't want to do that, so I tried some more educated guesswork. Can we guess the value of a without doing 3-digit multiplication? Well, the distance from 625M to C is 24,485,225. so, roughly how many 25ks can we add to that? A bit under 1000. In the 5 digit multiplication by hand, we have to roughly add 1000 of 25ks to get to C, meaning we need 2 of a bit under 500*25k, so I guess that a is probably 4 or 5. It could also be 6, but as we need a bit under 1000, I doubt that it's 6. So, we do 1 multiplication of a relatively simple 3 digit number, 255, then add 4 0s at the end to get 650,25 0,000. Well, that's actually kinda close. I'm now pretty certain that a is probably 4. I will assume this for now. So C is, in my eyes, almost certainly 254b5. Assuming a is indeed 4, we simply have to guess what b is. I have reduced the number of squarings I need to do if a is 4. I only need to do 4 squares now. But can we do more guesswork? Let's try to just look at 3 digit squares of 4b5. We can rule out those which do not end in 225, since the addition of the thousandth's and ten thousandth's digits will not affect the hundreth's digit and below. 415 * 415 = 172225, 435*435 = 189225, 465*465 = 216225, 485 * 485 = 235225. These are the only ones that match. So, most likely, b is 1, 3, 6, or 8. If that's true, then using binary search, I only need 2 5-digit square calculations to find the solution! If I want to double check though, I need 3. I am out of ideas now, so time to bite the bullet... Between b = 3 and b = 6, I first test 25465 because I expect the solution to be on the higher end of the range. It's 648,466,225. Well, if my logic and guesswork has been right so far, then the solution *should* be 25485, so... Well I'll be damned. 25485 is actually the solution. I used a calculator to double check after confirming it. This was rather pointless in the modern day thanks to calculators but ngl it felt pretty good when I confirmed my answer.
@Mk-rc7mj
4 ай бұрын
absolute mad lad
@thesciemathist6035
4 ай бұрын
I agree@@Mk-rc7mj Absolute Mad lad. Wish I had enough mental fortitide to stick with such a daunting calculation.
@icedo1013
4 ай бұрын
Being able to dissect the mechanisms of math and numbers and exercising your analytical abilities is *NEVER* pointless. Understanding why you're doing what you're doing is key to truly learning. You've obtained information by this practice that anybody else will miss because they skipped right to the end. Be proud that you wield the power to discern truths and deduce reason. This was very well done.
@lucahermann3040
4 ай бұрын
You could have divided by 25 and 9 first to deal with a smaller number. It's obviously divisible by 5, so first you divide by 25 and get 25979409. Then you check for divisibility by 3. 2+5+9+7+9+4+0+9=45 is divisible by 3, so 25979409 is also divisible by 3. So you divide by 9 and get 2886601. Since this number is between 1600²=2560000 and 1700²=2890000, you're looking for a number 1600+x with x
@owennewo14
4 ай бұрын
Leaving you on 69 likes coz that's dedication is nice
@JayTemple
5 ай бұрын
A is 1 less than 10,000^2. No square ends in 3, so B is out. D is divisible by 5 but not 25. That left C and E. The digits of D add up to 39, so it's divisible by 3 but not 9. Therefore, it's C.
@Noobish_Monk
4 ай бұрын
D is also divisible by 5 but not 25
@japanpanda2179
4 ай бұрын
E was actually the first I eliminated, in a much simpler way. It has 3 zeroes at the end, which means it's divisible by 1000, but not by 10,000. This means either 2 or 5 is in its prime factorization 3 times, so it can't be square.
@Mewhenthewhenthe-x7j
4 ай бұрын
@@Noobish_Monkyes. To find out if a number is divisible by 25 you can calculate mod 100. This is the same as looking at the last 2 digits and deciding if that is a multiple of 25. 35 is not a product of 25, so therefore the entire number is not divisible by 25.
@ethanlarge3572
4 ай бұрын
Easiest way to eliminate E: a square cannot have an odd number of trailing zeroes.
@goncalofreitas2094
4 ай бұрын
In the last one I think you meant "the digits of E", but very nice
@Near_Void
5 ай бұрын
I ruled out A B and E immediately and thought which one looked more like a square number was correct
@harry2.01
4 ай бұрын
Same here.
@AnthonyLOL..
4 ай бұрын
I ruled out A B and D immediately, then guessed C with the two left
@Platanov
4 ай бұрын
My reasoning was exactly the same! I looked at C and thought, well, there's a 49 and a 25 in there, it's gotta be a square number lol
@synexiasaturnds727yearsago7
4 ай бұрын
I picked the same as well, thought I was wrong because I mistyped the 6th digit
@Nikioko
4 ай бұрын
You can also immediately rule out d, because squares of numbers ending with 5 end with 25. Answer d ends on 35 and therefore is no square number.
@TrimutiusToo
5 ай бұрын
K×1000 can be a square but only if k is multiple of 10 so it will have 4 zeroes at the end not 3
5 ай бұрын
Actually it would be square when k=1000 then k*1000 = 1000*1000 which means that k*1000 is a square number on its own right (1000^2). And it alsa can be a square if you multiply it with an another square like (1000^2) * (10^2) or (1000^2)*(100^2) or even (1000^2)*(17^2) and so on. For example: k=4000 would be good. 4*1000*1000 = (2^2)*(1000^2) The rule: k must be (x^2 )*1000 then k*1000 is a square number. But then the number ends with 6 zeros at least not 3.
@kylejoecken2900
5 ай бұрын
@ this is not correct. A counterexample is k=10, where 1000k = 10000 = 100^2. The OP is right that since 1000 = 2*2*2*5*5*5, it must be that k = 2*5*n^2 for some natural number n.
@zanti4132
5 ай бұрын
Just factor option e as 1000 × K. K ends in 4, therefore is not divisible by 5. 1000 = 5³ × 2³, therefore has three multiples of 5. So when option e is broken down into its prime factors, it will have three 5s, i.e. an odd number of 5s, hence can't be a perfect square.
@user-dh8oi2mk4f
4 ай бұрын
@no, it’s k = n^2 * 10
@adw1z
4 ай бұрын
k*1000 = k* 2^3 * 5^3 to be a perfect square, need even powers of 2 and 5, which means an odd number of 2,5 factors in k (and even number of all other prime factors). But k must be a multiple of 10 as it must have at least one 2 and 5 to satisfy this, and clearly 987654 is not a multiple of 10, so clearly not a perfect square here. could have e.g. k = 2 * 5^3 * 7^2 = 49*250, and this would give a perfect square This is equivalent to saying k = 10n^2 for some integer n
@MrGeorge1896
5 ай бұрын
Perfect squares are either congruent to 0 or 1 (mod 4) so a) and d) are not possible. b) is not possible because of the last digit as explained in the video. Perfect squares can end with zeros but not with an odd number of zeros so the only remaining choice is c)
@rizzwan-42069
4 ай бұрын
I just eyeballed it and chose c.
@joelwillis2043
3 ай бұрын
@@rizzwan-42069 sad
@vylon1075
3 ай бұрын
Same thing I did. It will take just a glance if you know these.
@La_sagne
5 ай бұрын
25,000 ^2 should be 625,000,000 (if i remember 5th grade math correctly) and we got 24,485,225 remaining.. now every step between square numbers is 2 times the previous base number +1 so 25,001 squared must be 625,050,001. using that knowledge we can estimate.. lets go with 25,450.. that 625 million + 450 * 50,000 + 450 * 450 = 625,000,000 + 22,500,000 + 180,000 + 20,000 + 2,500 = 647,702,500... ok so were only 1.78 million away and we know most of that comes from the multiples 50,000 so lets go for 30 more.. that means 30*50,900 + 30*30 more so 1,500,000 + 27,000 + 900 more so a total of 649,230,400.. so were only 254,825 away... that looks a whole lot like 5 steps of a little over 50,000 each.. so my answer is 25,485^2 = 649,485,225.. wasnt really in my head and estimation isnt pretty.. but atleast it was without a calculator
@gutschke
4 ай бұрын
I made a similar argument and then also reasoned that the last digit has to be 5. Combine those two ideas, and it took me about half a dozen guesses to arrive at the correct result. That can easily be done with pencil and paper doing some long multiplication to narrow down the estimates. Shouldn't take more than maybe 10 minutes tops, but that could still be too much time for the purposes of taking a test.
@Ahmad-yi6d
5 ай бұрын
Square Root of A is 9999.99995 Square Root of B is 11105.55415 Square Root of C is "25485" Square Root of D is 31426.96295
@richardl6751
5 ай бұрын
Square Root of D is 26707.50896 Square Root of E is 31426.96295
@Bruh-bk6yo
5 ай бұрын
Therefore, every number is a perfect square! Huh, silly Oxford!
@nathan87
5 ай бұрын
I don't think you quite did (d) or (e) satisfactorily. For (d), you do not address the 50n terms. If we write out the multiplication we get (... + 100n^2 + 2*5*10n + 25) = (... + 100(n^2+n) + 25) so we see that 25 is the only term without a coefficient of at least 100. For (e), while it is true that sqrt(1000) is not an integer, that does not mean that sqrt(k)sqrt(1000) cannot be. For example, sqrt(10)sqrt(1000) = 100. The easiest way to do this is to remember that perfect squares can only end in an even number of 0s. This can be shown by considering that if x ends in n zeros such that x = X*10^n, x^2 = (X*10^n)^2 = X^2*10^(2n) which ends in 2n zeros.
@quantumgaming9180
4 ай бұрын
Your explanation for d) shouldn't take the 10's and 100's necessary equal to n. They are different in the problem
@nathan87
4 ай бұрын
@@quantumgaming9180 n is only for the 10s. That is, (... + 10n + 5)*(... + 10n + 5) = ... + 100n^2 + 100n + 25
@quantumgaming9180
4 ай бұрын
@@nathan87 ahhh I see now
@TheSuperSayahMan
4 ай бұрын
Looked at it, saw a number ended with 25 and guessed that was the answer. Skipped to the end and got it correct. That's what I'd have done in the exam and saved tb time for a question I knew how to solve as I'd have just wasted time on this one but I'm happy my first impulsive choice was correct.
@JediKnyghte
5 ай бұрын
With respect to d, a square of a multiple of 5 must have 25 as a factor.
@JayTemple
5 ай бұрын
And with respect to E, a multiple of 3 must have 9 as a factor. (It doesn't.)
@DanielMartin
4 ай бұрын
More than that: it must end in the two digits "25".
@jakit0556
3 ай бұрын
25485 because you can put numbers in your calc until you get the right digits and change them around until you get each digit, 1st one for 600mil, then fro the 40 mil then for the 9 mil and its relatively easy because you dont need to change the last digit as only something ending on 5 squared can produce 5 from the normal numbers
@urojony3177
5 ай бұрын
Here's how to do square root of 649,485,225 without calculator assuming you know that it's natural (even in your head if you have enough memory, which I don't have). Let x=649,485,225, y=sqrt(x) 1. x have 9 digits, so y have 5. 2. 25^2=625, 26^2=676, so first 2 digits of y are 25. 3. Last digit of y is 5. 4. Let y=10z+5, then y^2=(10z+5)^2=100 z (z+1)+25, so z(z+1)=6,494,852. 5. z(z+1)=2 (mod 25), 25 | (z-1)(z+2), z-1 and z+2 can't be both divisible by 5, so z=1 or z=-2 (mod 25). This means third and fourth digit have to be either 01, 26, 51, 76, 23, 48, 73 or 98. 6. 6,494,852=6-4+9-4+8-5+2=1 (mod 11), so z(z+1)=1 (mod 11), so z=3 or z=-4 (mod 11). 7. Let c, d be the third and fourth digit of z. We have -2+5-c+d=3 or -2+5-c+d=7 (mod 11), so d-c=0 or d-c=4 (mod 11). 8. Using 5 and 7 we have z=2526 or z=2548. 9. 6+4+9+4+8+5+2=38, so z(z+1) is not divisible by 3, so z≠2526. 10. Finally, y=25485.
@Ovoparity-jh6bt
5 ай бұрын
Take the last two digits and the middle 3 and your done
@sonicwaveinfinitymiddwelle8555
4 ай бұрын
@@Ovoparity-jh6bt ???
@cosumel
3 ай бұрын
I’m good at math, but this made my hair hurt
@neilgerace355
5 ай бұрын
Doing square roots without a calculator is similar to long division.
@wowyok4507
4 ай бұрын
nuh uh
@cyrusyeung8096
5 ай бұрын
Also, as I mentioned in 黑筆紅筆 , one can use "long division" to calculate square roots (search "square root by long division"). I commented again, so that English audience can also learn this technique.
@johnchestnut5340
5 ай бұрын
I saw that technique in a grammar school math book from the early 1900's. I have not seen it anywhere else. It uses '3' a lot.
@VeryGoodDeals
4 ай бұрын
Mentioned in what?!?!?!
@cyrusyeung8096
4 ай бұрын
@@VeryGoodDeals It is bprp's Chinese channel.
@mmn712jimmss
4 ай бұрын
In your head: A: 625m is the closest square that is smaller. ->25k + x is the number B: (25k+x)^2 =25k^2 +2×25k×x+x^2 / 2×25k×x = 649m-625m= 24m / 24m÷50k=480 C: last 2 digets are 25-> sqr is 5 ->> 25k+480+5= 25485
@SWLaw14
4 ай бұрын
Easy. Just put that numbers next to a square root and see which one reacts to it
@brettbishop2461
3 ай бұрын
So you can easily rule out A, B, E; so you're left with c and d. Then by counting you see D is only divisible by 3 once, so the answer is C Took about 12 seconds
@Biswajitjena.
3 ай бұрын
A 035 at the end will give 7 as last digit when divided by 5.. another way to rule out option d...
@chiehlilee9224
4 ай бұрын
It takes time but you can definitely find the square root of c without calculators by factoring the number. Knowing it’s a perfect square, you know all factoring prime numbers must be in pairs. So instead of 5, just divide it by 25. Then with other tricks, I.e. using the last digit + even #s are divisible by 2 (therefore 4) + if sum of all digits is divisible by 3 then the # is divisible by 3 (in this case 9) and so on until you have all the factoring primes. Then just separate the group and multiply all the primes of half of the pair together.
@TheBillBomb
4 ай бұрын
There is a long division method for doing square roots by hand. I learnt it in 1978, before calculators were cheap! Basically start at the big end and look at two digits at a time. Subtract the largest square , noting the digit. Bring down the next two digits. Need to know all the squares of two digit numbers upto 43.
@yurenchu
4 ай бұрын
"Basically start at the big end and look at two digits at a time." Don't forget to first add a leading zero when the number to be square-rooted has an odd number of digits. 06 49 48 52 25 || *2 5 4 8 5* 04 = *2* * *2* ---- 02 49 02 25 = ( 2*20 + *5* ) * *5* --------- 24 48 20 16 = ( 2*250 + *4* ) * *4* --------- 4 32 52 4 07 04 = ( 2*2540 + *8* ) * *8* ------------ 25 48 25 25 48 25 = ( 2*25480 + *5* ) * *5* -------------- 0 ==> 649,485,225 = (25,485)^2
@tomasstana5423
5 ай бұрын
The explanation why square ending in 5 must end in 25 is wrong/incomplete. When you have (... + 10n + 5) * (... + 10n + 5), you don't just get 100n^2, but you also get 50*n, but two times so it is also multiple of 100, and therefore does not affect the first two digits( and multiples of anything else are multiples of 100 trivially)
@sodabutnofizz1294
4 ай бұрын
Easily (c) 100,000,000 is already a perfect square and a number just 1 less than it is less likely to be a square so option (a) eliminated A perfect square cannot end with 3 so (b) eliminated If theres a 5 at the end and the number is a perfect square than the digit preceding 5 must be 2 or else it is not a perfect square (You can have squares of numbers with 5 at units place end with 25 ALWAYS but no other digit can ever come there) And in (e), well, it ends with 3 zeroes, cannot be a perfect square as to be one it must end with even zeroes.
@SujayDas-sd44sd19
4 ай бұрын
(C) can be figured out using long division method for finding square roots.
@tgg7525
3 ай бұрын
Mod 4, a) and d) cannot happen because 3 is not a quadratic residue. Mod 5, b) cannot happen because 3 is not a quadratic residue. Looking at modulo 7 we easily convince ourselves that e) is equal to 6 which is not a quadratic residue modulo 7. The correct answer is therefore c).
@sterlingbuck967
4 ай бұрын
Another fun thing about squares is that the digital root of a square will only ever be 1, 4, 7, or 9
@Ard5000
4 ай бұрын
my solution before seeing the video ( btw i am in grade 11th in india) e.) cannot be a sol as it has 3 zeros a and b.) cannot be as 333 and 999 must belong to 11 and we don't see a pattern as with 11 such as 12321 and much more d.) as 035 as last digits do have 7 and 35 but there is no root as of till i know have 35 at end c.) 225 is a sq of 15 i have yet to see the ans so don't see my ans and say i am correct or not i maybe wrong
@keshavgoyal366
4 ай бұрын
I have a doubr in the e option Suppose the no is a multiple of 1000 1000x Root of 1000x is under root 1000 Multiplied by root of x If x is also a multiple of one thousand by a perfect square then it will equate to something like a 1000^1/2 × 1000^1/2 which is equal to 1000a which is a natural number
@aniruddhaganbote9327
3 ай бұрын
You can do it by eliminating option like no square has 3, or 000, or 35 at the last, and option a can be easily checked so the best shot is option c
@HenrikMyrhaug
5 ай бұрын
I thought A, B and E were obvious since A is a perfect square minus 1, B is divisible by 3, but not 9, and E is divisible by 1000, which is not perfect square. It should have been obvious to me D is not divisible by 25, since it doesn't end with 00, 25, 50 or 75.
@Mewhenthewhenthe-x7j
4 ай бұрын
We’re calculating quadratic residue with this one🗣️🗣️🗣️
@Evisceratio
4 ай бұрын
А isn't correct, because 99 999 999 = 9 x 11 111 111 = 9 x 11x 101x 10001, 101 and 10001 don't divide 11. B isn't correct, because in 123 333 333 sum of digits (24) divides 3, but doesn't divide 9, so number divides 3 and doesn't divide 9 D isn't correct, because it ends at 35, square of number ending on 5 ends on 25. E isn't correct, because there are three zeros in its end So by exclusion - C
@jedinaj
5 ай бұрын
I saw 225 at the end and guessed C 😅
@yiutungwong315
5 ай бұрын
Alright 👌👍
@cauthrim4298
4 ай бұрын
A) Divisible by 11, but only once B) Perfect square can't end with 3 C) The answer by elimination D) Ends with 5, but not 25 E) Odd number of zeroes
@woshdndndj2103
4 ай бұрын
I did the same but for A 99,999,999 is 1 difference from 100,000,000=10000^2 so it cannot be a square
@Inequalito
5 ай бұрын
Amazing video! I will be trying to analyse big numbers like this now, very interesting!!
@cyrusyeung8096
5 ай бұрын
I have just watched the video in 黑筆紅筆 2 hours ago, and I am now going to watch again.
@bprpmathbasics
5 ай бұрын
Thanks, haha!
@pelayomedina2174
4 ай бұрын
c) Ez Basically use properties of 3 and 9 multiples
@inboundbark
5 ай бұрын
All odd perfect squares are congruent to 1 mod 8 so you just need to check the last 3 digits mod 8 (e.g 225 mod 8) for the first four options.
@Quantris
3 ай бұрын
you mean last 3 digits. good fact to remember!
@inboundbark
3 ай бұрын
@@Quantris yeah you're right! When you see the numbers represented as a sum of powers on a daily basis you tend to think of the coefficients of the lower powers as first!
@sonicwaveinfinitymiddwelle8555
4 ай бұрын
just use long division, it is faster and takes less effort
@A_Wandering_Fanatic
4 ай бұрын
The square of a number having 5 as its unit digit always has 25 as its last digits. For example, the square of 15 is (1*2)25 is 225, for 25 it's (2*3)25 i.e. 625.
@Laggron93
5 ай бұрын
e is not because it's a multiple of 10^3 but not 10^4. d is not because it's a multiple of 5, which means it has to be a multiple of 25, which it is not. b is not, because no perfect square ends with 3. a is 9 x 11.111.111 = 9 x 11 x 10.101.010, but 10.101.010 is not a multiple of 11 (odd ranked digits add to 0, even ranked add to 4), which means 99.999.999 is a multiple of 11 but not 11^2, thus not a square. Which leaves us with c: 649.485.225.
@МаксимАндреев-щ7б
5 ай бұрын
a) n=3 (mod 4) but square must be = 0 or 1 (mod 4) b) square can't end on 3 d) n=3 (mod 4) e) square must end on even number of zeros So the answer is c
@ayushrudra8600
5 ай бұрын
To look between c and d you can say that squares are either 0 or 1 mod 4 and d is 3 mod 4 therefore its c
@bobh6728
5 ай бұрын
Didn’t know that squares are all either 0 or 1 mod 4.
@nathan87
5 ай бұрын
@@bobh6728 It is very easy to show, consider odd and even integers separately :)
@bobh6728
5 ай бұрын
@@nathan87 After I posted I looked it up. Just had never heard of that before. Even at my age, I can learn something new.
@bloodyadaku
4 ай бұрын
If you want to find the square root of a number that you know has an integer square root, your best bet is to factorize it and then match the factors in pairs. Because it is a square number, every factor will have a pair. Then just multiply one factor from each pair together and you will have the answer. Of course, factorizing a number is not always easy especially when the factors are large primes. But you can generally be sure that this may not be true because the set of numbers with large prime factors is significantly small as the number gets large.
@zoro8800
5 ай бұрын
guessing option c (before watching the video)
@fifiwoof1969
5 ай бұрын
Yeah pretty easy.
@quigonkenny
5 ай бұрын
Regarding C, for any number that ends in 5, its square will be of the format 100x(x+1)+25, where x is the digits before the trailing 5. For example, take 225². 22•23 = 506, so 225² will be 50625. Don't know any trick on how to then get the square root from the final square, though, short of using the usual square root estimation methods on the digits before the 25 to get close enough to tell from that final digit (which will be either 0, 2, or 6).
@flameofthephoenix8395
6 күн бұрын
Hm, doesn't seem too tricky, to be perfectly lazy I'll just break it all up into its factors, an easy factor for 99,999,999 is 10,001 making it 10,001*9,999 and then 10,001*101*11*9, which is frankly ridiculous that it had that many repeat numbers, but we'll of course keep going, 11 is prime so it can only be solved by a second 11 we'll keep that in mind but remove it for now, nine is a perfect square of 3*3 so it will be removed without being noted down, we know that none of the 10...01 numbers are multiples of three because ten divided by 3 is 3 with a remainder of one meaning that the tenths place digit in a number will only add itself multiplied by one to the modulo this means that no matter what position the two ones are at in a number they will still only add 2 to the modulo operation, none of them are multiples of 5 either because all multiples of 5 must end in a 5 or a 0, to determine if they are multiples of 7 we can convert this problem into X = number of 0s between the two ones, and 0=(3^(X+1)+1)%7 which is ever so slightly easier I think, for 101 X = 1, 3=(3^(X+1)+1)%7 so 101 is not divisible by 7, for 10,001 X is 3, 5=(3^(X+1)+1)%7 which means that 10,001 is also not divisible by 7, the next prime is 11 which means we have to break our numbers up into pairs of 2 digits to determine the modulo and therefore whether they are divisible but also only have to add these pairs together in order to find it, 101%11 is 2 and since 11 squared is 121 which is greater than 101 this means that 101 is prime which means that 99,999,999 is not a perfect square because 10,001 does not contain the factors 101 and 11 which we can confirm because 101*11 is 1,111 which can easily be shown unable to divide into 10,001 because 1,111 times nine is 9,999 which is two away from 10,001. This has proven the first of the five answers incorrect since 10,001 would have to contain a match for 101 and 11 for them to be parts of a matching set which is required for a perfect square. Next there is 123,333,333 which will likely prove significantly more difficult, for this one I'll use a more iterative approach, first I'll find a good perfect square starting point, 100,000,000, the sqrt of this is 10,000 to get from this square to the next square we need to add 20,001 which means a good start is to divide 23,333,333 by 20,001, we don't need to because we can already see a good fit is 1000, this will bring us from a sqrt of 10,000 to 11,000 and the square gets added to by the sum of 20,001, 20,003, 20,005... ... 21,995, 21,997, 21,999 which we can find by multiplying (20,001+21,999)/2 by 500 or 21,000,000 which added to the square becomes 121,000,000, to get from this to the next square we need to add 22,001, so now let's just divide 2,333,333 by 22,001, the best fit appears to be 100 so we'll now do the same thing as before making the square become 122,105,050 and the sqrt become 11,100, which is starting to make me think it might just be 11,111 so I'll just square that to see how that goes, 11,111 * 11,111 = 123,454,321 which is pretty close, but not quite there so I'll divide the difference which is -120,988 by 22,223 which we'll just estimate to 6 which makes it become a sqrt of 11,105 which has a square of 123,321,025 the next square is 123,343,236, 123,333,333 is between these two squares meaning it is not a perfect square. 649,485,225, this one could be very easy to disprove/prove, just divide it by 25 first 6,494,852.25 > 25,979,409 which is a whole number meaning that all we've done is simplify it a little bit, next I'll check if it is divisible by 9 by adding each digit 2+5+9+7+9+4+0+9=18, I removed the nines which contribute nothing and since 18 is a multiple of 9 that means we can commit to the division and divide the whole thing by 9 making it become 2,886,601, we can check divisibility by 9 again just in case 2+8+8+6+6+1=31 3+1=4 which is not a multiple of 9 meaning that we'll have to move on from that 9, we could try 11 which means 1+66+88+2=3 given that 66 and 88 can be removed since they are multiples of 11, we could try 7, to make it easy we can group the digits into bundles of 6 because 999.999 is divisible by 7 which quickly cuts back the modulo operation to 886,603%7 next we'll just multiply the left most digit by 3 and add it to the one on the right repeatedly getting a modulo of 4 meaning it is not divisible by 7 either. After this I'll just lazily approximate its square root by adding each pair of digits and halving their place value making it become 3721, next I'll use Newton's method to refine my approximation to 2249, this shows that my approximation is devastatingly quite far from what it needs to be, and I just realized the method I used earlier to find the sqrt of answer B was in fact Newton's method but unrefined, intuition is going to go ahead and let me skip straight to 1500 which seems like a good guess as to the geometric mean of 1000 and 2249 and thus hopefully closer to the square 2,886,601, now I'll make another educated guess to skip to 1600, the next iteration brings it to ~1700 which is rounded to the nearest hundred, the next iteration brings it to a range of 1698 to 1700 1698 squared is 2,877,394, the next square is 2,880,791, and after that is 2,884,190, this shows I clearly failed at Newtonian method because the high end of the range is less than the square, the next square after that is 2,887,992, huh. It's supposed to be 1699. That's odd. Ah, whoops! It seems I failed at my initial squaring of 1698 which of course threw off all later squares due to how I was finding them, 1698*1698 should be 1000*1698+600*1698+90*1698+8*1698 or 1000*1698+700*1698-2*1698 or 1698000+7*169800-2*1698 or 1698000+1188600-3396 or 2,883,204. That took much longer than expected! It's almost midnight now, you stole my day away!
@SuryaBudimansyah
5 ай бұрын
Why do they have to say "Exactly one" instead of just one?
@briankelly5828
5 ай бұрын
Because they're mathematicians.
@emurphy42
5 ай бұрын
@@briankelly5828Because "one" could be interpreted as either "exactly one" or "at least one", depending on context. (Say someone sees a herd of cows, gets a better look at the nearest one, and says "Look, one has brown spots", without meaning to rule out the possibility that one or more of the others do too,)
@tambuwalmathsclass
4 ай бұрын
Can we try linearization?
@Xzw557blown_up
4 ай бұрын
Much longer method, you could do a prime factorization of each number, if the powers of all the prime factors are even, then the number is a square number. For example: 7056 has prime factors of 2x2x2x2x3x3x7x7, which can be written as 2^4 x 3^2 x 7^2. all of the prime factors are in powers which are multiples of 2. 7056 is 84^2
@ferusskywalker9167
4 ай бұрын
That's what I would have done if given this problem
@wealthychef
5 ай бұрын
At 4:15, you appear to be saying that you cannot multiple two irrational numbers and get a rational number. Is that a proven fact or just a conjecture? :-)
@bprpmathbasics
5 ай бұрын
I didn’t make myself clear because it was a bit too obvious in my head. It’s possible to have a product of two irrationals to be a ration. Such as sqrt(2)*sqrt(18) But I knew sqrt(987654)*sqrt(1000) for sure cannot be rational since the first one doesn’t have a factor of 10.
@wealthychef
4 ай бұрын
@@bprpmathbasics Thanks I think I get it. It looks like you are saying that there is a "dangling 10" in the sqrt(1000) term that has to be "married" somehow to become rational. In sqrt(2) * sqrt(18) I can see you pulled out the 2 from the 18 and "married" it to the other 2, is how I think of it. I like shortcuts, I'm too lazy to even get out a calculator most of the time, always like to try to solve things in my head. LOL
@mahdimozafari2663
3 ай бұрын
C stands for correct
@hamzaiqbal7178
4 ай бұрын
C looks the most correct Damn I did it. It literally stands out from the others due to it's end
@ZantierTasa
4 ай бұрын
I managed to narrow it down to (a) or (c) within a minute, but then struggled to prove that (a) isn't square. Here's 2 alternative (very similar to the video) ways to explain (a) ------- 1 ------- The gaps between square numbers increase as the numbers get bigger. Square numbers: 0, 1, 4, 9, 16, 25, .... Gaps: 1, 3, 5, 7, 9, .... 100,000,000 is square, and 99,999,999 is too close to also be square ------- 2 ------- More directly, 100,000,000 is (10⁴)², and the square number below it is (10⁴ - 1)² = 99,980,001. There can't be any squares in between.
@thewoodfamily6511
5 ай бұрын
Team C 👇
@katherinyu9346
3 ай бұрын
Your explanation for (d) leaves out some important details. You have (10n+5)^2 = 100n^2 + 100n + 25. It's not obvious a priori that the middle term (2*10n*5) is a multiple of 100, which makes it not interfere with the 25, so that's a nontrivial detail. Also, for (e), you need to say that $k$ is not divisible by 10 (we know this because otherwise, you would have an additional zero at the end).
@SmugBora
4 ай бұрын
I guessed C and checked my calculator and I was right lol I looked at it for a while and noticed it ends at 25. *Just like every number that ends with 5, the number ends in 25* Ex. 5^2 = 25 15^2 = 225 25^2 = 625 105^2 = 11025
@grishnakh
5 ай бұрын
a. =10^8-1; b. only 3^1, d. only 5^1, e. contain 10^3, only c is possible. Assume c= (5x)^2, clearly 5 digit, 26^2 > 25.5^2 > 25^2+2*25*0.5 = 650 > 649 > 25^2+2*24*0.5(+0.25) (= 25.4^2) So the final answer will be 254y5, and will contain factor 3 y can only be 2,5,8. Assuming the number is 100T+25/55/85, only (100T+85)^2 can end with 225 (25 will be 625, 55 will be 025). Therefore y=8, c=25485^2
@Nebula_ya
3 ай бұрын
My attempt before watching: a) 1000² -1, 1 less than a square so not square. b) Has exactly ome factor of 3, 3 is not a perfect square so this number isn't. c) I think this is square by process of elimination. d) Similar to b, has exactly 1 factor of 5, 5 is not perfect so neither is this number. e) Has exactly 3 factors of 5, odd numeber of factors so not square.
@JayNaCall
3 ай бұрын
My own was just by elimination... A. This looks like a perfect square because 3*3=9 but 33²=1089(if my calculations is correct... Just too lazy to grab pen and paper) so not a B. There is no such squared number ends with 3... 0=0,1=1,2=4,3=9,4=16,5=25,6=36,7=49,8=64,9=81 So not B D. D on the other hand ends with 5 so it could be any number that ends with 5 squared. But any number ends with 5 squared the result must end with 25. Like 15²=625. Therefore it's not D E. On the other hand might look like perfect square because it has 0 at the end. But if you square any numbers that ends with 0 the number of zeros always doubles. (100²=10,000 note from having two zero's it now had four zeros. 15,000,000²=625,000,000,000,000. From having 6 zeros now it has 12 zero's) The letter E ends with 3 consecutive zeros and 3 isn't divisible by 2 therefore it isn't perfect square. The answer is C... Why? Since it ends with 25, therefore if we find the square root of it it must end with 5. There fore it's C... Total guess before watching this video.... I solved by looking at the thumbnail...😊😊
@TrezkotTheFool
4 ай бұрын
N=649485225>(25000)² because 25²=625. 251²=60301 so N>(25100)². (25000+100k)² =6250000+5000000k+k²*10000 The important part is the middle term (because the last term add just a little to the result), if we study this part we can conclude that k=5 and k=4 give us an upper and lower bound of N respectivetly. So N>(25400)². We repeat this process with (25400+10k)²=645160000+508000k+k²*100 and we conclude that k=9 and k=8 are an upper and lower bound for N. So N>(25480)². Finally, N is a multiple of 5, so (25485)²=N.
@alcesmir
4 ай бұрын
Verifying C wasn't too hard with some slightly clever brutish work. Note that it's a multiple of 25, and do the tedious division 649485225/25 = 25979409 Note that the remains is divisible by 9, again do the division 25979409/9 = 2886601 = x^2 Note that x can only be a square of something ending in 1 or 9 Now it gets a bit more annoying, but note that this is x^2 = 2886600 + 1 -> x^2 - 1 = (x - 1)(x + 1) = 2886600 So x^2 ≈ 2886600 -> (x/10)^2 ≈ 28866 Now some guessing, 16*16 is 256, 17*17 is 289. This latter one seems very promising, but it's a tad too high. Given that x ends in 1 or 9, guess that x is 1699. Verifying 1699*1699 = (1700-1)^2 = 2890000 - 3400 + 1 = 2886601 !!! So 649485225 = 25*9*1699^2 -> sqrt(649485225) = 5*3*1699
@Siddhartha.Chatterjee
4 ай бұрын
Not B: because no square number ends in 2,3,7 and 8 Not E: because a square numbers ends in no or even number of zeroes (k*10^n)^2 = k^2*10^2n = k^2 * 100^n Not D: (had to hit and trial) because if a square number is to end with 5, it should end with 25 (examples: 25, 225, 625 etc.) (10k+5)^2 = 100k^2 + 100k + 25 = 100(k^2+k) + 25 Not A: (had to hit and trial) because if a square number, say N, is to end with 9, then N//10 is divisible by 4 (examples: 9,49,169,289,529,729 etc.) (10k+3)^2 = 100k^2 + 60k + 9 = 4(25k^2+15k) + 9 = 4h+9, and (10k+7)^2 = 100k^2 + 140k + 49 = 4(25k^2 + 35k + 10) + 9 = 4h+9, where k is any integer By elimination, it is C
@user-ll1ow4xf8n
4 ай бұрын
For c) I divided by 5^2 to get 649,485,225 = 5^2 * (25,979,409) Those digits sum to 45 so we can divide by 3^2 to get 5^2 * 3^2 * (2,886,601) None of my tricks seem to work so I consider x^2 = 01 mod 100, this implies x = 1, 49, 51, or 99 mod 100 This reduces a lot of possibilities so I started guessing. 2000^2 is way too high, so were 1900^2 and 1800^2, but 1700^2 = 2,890,000 which is barely higher, so I guessed x = 1699 and indeed that squares correctly. Thus sqrt(649,485,225) = 3*5*1699 = 25,485 A decent bit of arithmetic, but not too bad
@Wolf_Avatar
4 ай бұрын
Ok, calculating the square root by hand: 649,485,225 is obviously divisible by 25. Dividing this out gives 25,979,409. 25,979,409 is divisible by 9. Dividing this out gives 2,886,601. 2,886,601 has no obvious prime factors, so... I know there's a way to do square roots by hand, but I can never remember it. So, let's use Newton's method. Removing 4 digits, we've got 288. That's almost 289, which is 17 squared, so our first estimate is 1700. 2,886,601 divided by 1700 is 1698 plus a fractional bit I didn't calculate because we're assuming a whole number anyway. Which means... we get 1699 after one step? Quickly square 1699 to make sure. Yup, it's 2,886,601. So.. 1699*5*3 = 25,485. So that's the answer.
@ZvonimirZelenika
4 ай бұрын
a/b/d/e can be eliminated very quickly - 100.000.000 is square of 10.000, so one number less can not be, b) square can not end on digit 3 and also d) any square of number ending with 5 must end with 25, e) would need to have even number of 0s at the end. So if one IS square - then it's c)
@richardfarrer5616
4 ай бұрын
The square has 9 digits so it's root will have 5 digits. We know 25x25 = 625 so 25,000^2 = 625,000,000 is a good start. So (25,000 +X)^2 = 625,000,000 + 50,000X + X^2 = 649,485,225. Ignore the X^2 to get X is approximately (649,485,225 - 625,000,000)/50,000 which is about 24,000,000/50,000 = 480. We also know the root ends in 5, so the first attempt is 25,485, which is the correct answer
@ALeafOnTheWind42
4 ай бұрын
A more thorough explanation on eliminating d, that comes with a trick for squaring any integer that ends in 5. Any perfect square that ends in 5 must have a square root that ends in 5. Suppose d is a perfect square, and let n = sqrt(d). Because n ends in 5, there exists natural number k such that n=10k+5. We know that d= n²= (10k+5)² = 100k² + 100k + 25 = 100k(k+1)+25. Note that because k is a natural number, 100k(k+1) ends in 2 zeroes, and thus 100k(k+1)+25 ends in 25. But notice that this gives you a trick for squaring any integer that ends in 5. Simply take all the duties digits that precede the 5, then multiply that number by 1 more than itself, and stick a 25 at the end. For example, 35², do 3*4=12 and append a 25: 1225. 105²? 11025
@Utesfan100
4 ай бұрын
We know it is an integer and the last digit is a 5. Indeed, the ending 225 requires an ending of 15, 35, 65 or 85. It is about 649*10^6, which is half way between 25*10^3 an 26*10^3. Thus we can use Newton's method with an initial guess of 25505. x/25505 is 25472.8..., so a better guess is near the mean 25489, which is near 25485. x/25485 is 25485, so this is indeed the root. Indeed we, could have started with 25005. x/25005 is 25974.0... so a better guess is near the mean 25489.5, which is near 25485. This also shows anything between 25000 and 26000 gives a second guess near 25489, which would suggest 25485. If we had made a worse approximation and not ruled out 25495, x/25495 gives 25475.0, with a mean near 25485.
@saltdestroyer
3 ай бұрын
i mean if you know a number is a perfect square there are ways to approximate its sqrt. 649 is the next number betweeen 25 and 26^2, so the number starts with 25. The number ends with 225, which is 15^2. So we have 25_15 right now. the middle number is 485, which is 22^2 + 1. So split the 1 off, then make the end 1225, which is exactly 35^2. Double the middle sqrt, add it in, to get 25,44_. add in the 35, you get 25,475, which is my approximation
@Engy_Wuck
4 ай бұрын
A is divisible by 3², (99,999,999 = 9 x 11,111,111) and 11 (--> 9 x 11 x 1,010,101), but not 11² . so not a square. B is divisible by 3, but not by 3², so not a square C in divisible by 3² and 5², so a candidate (next is 1699², so not easily checkable) D is divisible by 5, but not 5², so not a square E is divisible by 3, but not 3², so not a square. You don't need fancy rules, only low-grade school math 🙂
@ethos8863
4 ай бұрын
For the first one, perfect squares must be 2n+1 apart. For example 12 squared and 13 squared are 12+12+1 apart. This is because 13^2 = (12+1)(12+1) = 12^2+12+12+1 when foiled. So we know that numbers very close to a perfect square cannot be perfect squares themselves. So since A is very close to (10^4)^2 it must be at least 2*10^4+1 away. B doesn't look like a square. E isn't a perfect square because it's a multiple of 1000 and not 10000 and no number square to 1000 so it's also not a square. So it's either C or D. This is as far as my cursory glance gets me. B doesn't look like it follows the digit rules for squares and it's also a multiple of 3 and not nine. squares have a requirement that every factor has a pair.
@UltraMC3
4 ай бұрын
the square root of 649 is roughly 25.48: 649 -625 (25^2) = 24 24/50 = 0.48 so we know that 649,000,000 is roughly 25480^2 Because 649,485,225 is greater than 649,000,000, so the answer is either going to be 25485 or 25495, because it ends in 5 So now we focus on the last 3 digits of both to figure out which one ends in 225. Because 400 * 5 is 200, we can reduce that to the last 2. 85^2 = 6400 + 800 + 25 = 7225 95^2 = 8100 + 900 + 25 = 9025 This means that the square root of 649,485,225 is 25485
@TheMathManProfundities
2 ай бұрын
Your explanation for the final option is not sufficient. Just because √1000 is not an integer does not mean that √(1000k) isn't. For example, k=40, √(1000k)=√(40,000)=200.
@philippenachtergal6077
3 ай бұрын
hum. How to do that quickly. Obviously it's not b because it ends with a 3. Cannot be d either because (10n + 5)^2 must end by 25 Cannot be e either as a square number ending by 0 should have an even number of 0s at the end (if a number j is written as k * 2^n * 5^n and k is not a multiple of 10 then j^2 is of the form k^2 * 10^(2n) where k^2 is not a multiple of 10) Cannot be a because a = 100.000.000 - 1 and 100.000.000 is a square number (the only two consecutive square number are 0 and 1) So by elimination, it is c
@ethanlarge3572
4 ай бұрын
Before watching: D: no, because the square of a number ending in 5 will end with 25, and a square ending in 5 must be the square of a number ending in 5. B: no, because a number ending with 7 or 3 cannot be a square. The square of a number will have the same final digit as the square of its complement (it’s quite cool; the final digit sequence goes 1-4-9-6-5-6-9-4-1-0). I haven’t done a proof of this but it should be pretty simple. A: no, because the square of 10,000 is too big (by 1) and the square of 9,999 is too small. E: no, because a square cannot end with an odd number of zero digits. Again, I haven’t proven this but it’s clear it must be true. Thus, assuming exactly one correct answer, C must be a square (it would have been the prime candidate for a guess if necessary, because it ends in a square). QED
@chriscobe1990
4 ай бұрын
very wonderful video, well done.Only a consideration: I agree with the fact that the answer D) cannot be a perfect square, but I disagree with the explanation because you said “we have square of k•1000 , I don’t know K but 1000 doesn’t give a perfect solution, so it cannot be this”. Actually the solution depends on K, for example the number 160.000 is a multiple of 1000 with K =160. Following your reasoning it should not be a perfect square while we know square of 160.000 is 400.
@WilliamCacilhas
4 ай бұрын
I knew A was out because no whole number multiplied by itself could give that number. There is a pattern that arises when multiplying 9’s: 9*9= 81 99*99 = 9801 999*999 = 998 001 9999*9999 = 99 980 001 As can be seen, there is always an 8 and a 1. The 8 and 1 get separated by 0’s equal to the number of 9’s that follow the first 9 in the number being squared. You also add 9’s to the front equal to the number of 9’s that follow the first 9 (for example 9999 is 9 followed by 3 more 9’s). B, D and E all looked wronged to me as well although I had no reason for why. C looked like the most likely option. My intuition was correct. I wish my intuition was correct like this more often but unfortunately I’m wrong more often than I am right.
@livenlife453
4 ай бұрын
My method for solving: Not A because of how close it is to 10^8 Not B because it is divisible by 3, but not 9, which is not possible for a perfect square. (Add digits trick) Not D because a divisible by 5 number must be divisible by 25, which that number clearly isn’t. Last 2 must be 00,25,50, or 75. Not E because of same divisible by 3 but not 9 trick as answer B.
@Loaderdani
4 ай бұрын
649 485 225 / 25 = 25 979 409 25 979 409 / 9 = 2 886 601 2 886 601 is not divisible by 81, so I don’t know how to get further by hand. I could brute force it, but I may have to try hundreds of numbers to find the next factor, which I know will end in 9.
@CrYou575
4 ай бұрын
649,485,225 obviously has a square root greater than 25,000 but less than 26,000. One step of a Newton approximation should give the next digit as 4. So 254x5, because we know the root must end in 5. The square must be a multiple of 3 (because the original number is a multiple of 9), so x = 2,5 8. Squares of numbers ending 25 end in 625, ending in in 55 end in 025 and ending in 85 end in 225. Therefore x is 8, giving 25485 as the root.
@tylerduncan5908
4 ай бұрын
Before watching, my observations are: A is incorrect because 99,999,999+1=10⁸. Consecutive integers with absolute value > 1 are never both squares. B is incorrect because sqaures can't end in 3, since there are no 1 digit numbers whose square ends in 3. E is incorrect because it is of the form n×10³. Since the factor 10 is raised to an odd power, it can't be square. This leaves us with C and D, which, if perfect sqaures, the last 2 digits must be of the form 5+k×10 We can check these cases quicly to find that no squares end in 035 making the answer C
@user-ll1ow4xf8n
4 ай бұрын
A) I divided by 3^2 = 9, the result was 11 mod 100 which is impossible for a square number B) This is 3 mod 10 which is impossible for a square number D) This is 5 mod 10, but not divisible by 25, which is impossible for a square number E) I divided it by 10^2 = 10, which is 0 mod 10, but it is not divisible by 100, which is again impossible
@joshuaanoruo973
4 ай бұрын
Eliminated A because: 99,999,999 is 100,000,000 - 1 and it's a difference of 2 squares Eliminated B because: No squares end with 3 Eliminated E because: No squares end with an odd number of zeroes It was between C and D, then I after that the square of any number that ends with 5 must end with 25. That eliminates D and puts C as my answer
@Misteribel
4 ай бұрын
On (A), a repdigit can never be a perfect power (Bugeaud and Mignotte, 1999). It's trivial to see for any digit, except 1, which was proved in that paper. (B) no ending in three (C) candidate (D) nope, 7x5 is not 5x5 (E) nope, squares ending in zeroes require even number of zeroes
@weeniethepoop2550
4 ай бұрын
Wow my intuition was correct. Not A because although 3² is 9, 33² is 1089 and not 99. Not B because you can't square to get a 3. C is my answer cause it looked like 5² Not D cause 35 is impossible to get from a square. Not E because it's 1000*something
@goncalofreitas2094
4 ай бұрын
The justification of E in the video is wrong, if for example k=10 then k*1000 is a perfect square, so the square root of k does matter For me the easiest way to see that E is not a perfect square is that it's divisible by 10^3, and so it's divisible by 5^3 But it's not divisible by 5^4 because divinding it by 10^3 = 2^3 * 5^3, the result is not divisible by 5 (ends in a 4) So 5 only appears an odd nb of times in its prime factorization, therefore it can't be a perfect square
@amadeus6834
4 ай бұрын
We can divide A by 9 and 11 (the second is a prime number), and we get 1010101, but its not divided by 11 again, because of its rule; 1-0+1-0+1-0+1 = 4, and 11|4 is false. Therefore, de original number is divisible by 11 once, not twice, so it can't be a square number. But, what we can se in the video, is more elegant.
@sterlingbuck967
4 ай бұрын
The thing about C is it mildly follows some lever of 7 principle with the 649 start because I want to say that is how 7^6 ends. Following the logic that E begins with 987 which is a Fibonacci Number I could imagine that the answer would be 17711^2 because that's the 22nd fibonacci number. I haven't used a calculator yet but I'll edit my response once I've figured it out. I was incorrect on so many levels lol
@VeteranVandal
4 ай бұрын
No square number ends in 3. No number ending in 5 has a square that doesn't end in 25. Every number ending in 000 cannot be a square. 10^8 = 100000000 and 9999² ends in 1. So only C makes sense. Oh, you used the exact same things I used, goddammit. :(
@MultiPaulinator
4 ай бұрын
I'm commenting before watching as encouraged. A and B are blatantly not it. E has an uneven number of trailing zeros which is impossible for a square. D ends in 035 and 35=5*7 making it at least highly unlikely to be a square. So, I checked the square root of C and got an integer.
@Nikioko
4 ай бұрын
Square numbers end on 1, 4, 25, 6, 9, or an even number of 0s. That rules out options b, d, and e. 100.000.000 is a perfect square, 10.000², and 99.999.999 is 1 less than that and therefore obviously not a square, which rules out option a. So, the remaining answer c must be the correct one.
@dontich
3 ай бұрын
my answer : a&b are clearly not perfect squares -- way too many repeated digits. d ends in 35, every PS with a 5 ends in 25 not 35. e ends in 000; each perfect square with zeros should have an even number. Therefore C.
@low-litlight3438
3 ай бұрын
Really easy problem tbh, but I’m a mathematician. You just need to check A, B, D, and E for divisibility by small primes and determine if the prime factors divide the number an even number of times: this doesn’t hold true for those 4 problems. Since we know one must be correct, we know it must be C.
@fernandojackson7207
4 ай бұрын
First one is 10,000^2 -1, so not a square. For b), no square ends in 3( nor 2 or 7) . Similarly , no square ends in 35. Last we can divide by 100 to 9876540.
@lucafreyq
3 ай бұрын
The argument for e doesn't make any sense to me. 10000 is also some k multiplied with 1000. Also √16=√8*2. There you could also say √2 isn't a whole number and we rule it out.
@FractalMannequin
4 ай бұрын
First is congruent to -1, hence 3 modulo 4, so it's not a square. Second's rightmost digit is 3, so it's not a square. Fourth is congruent to 3 modulo 4, so it's not a square. Fifth is divisible by 5³, but not 5⁴, and squares must have even powers in prime factorisation, so it's not a square. The only possible square is the third.
@SimonaDaRat
4 ай бұрын
I used my FEELING and i actually got it correct , im surprised
Пікірлер: 412