This is how University of Oxford asked a square number problem on its admission test
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- Опубликовано: 3 май 2024
- This question is from the 2023 University of Oxford Maths Admission Test. Try the problems here: www.maths.ox.a...
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By squaring 25,485 different numbers I have concluded that the square root of 649,485,225 is 25,485
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".
@@pietergeerkens6324I don't understand that Cuz I'm 12 years old but it looks cool😮
@@ZDTF If you have a specific question I will have a go at explanation.
@@pietergeerkens6324 what is a centit
Damn, without thinking too much about it i thought of ~25000 (+5). Wasn't too far off!
I guessed C without even thinking about it, put me in Oxford
sanme
same
Coming to a conclusion without actually thinking about it sounds more like Harvard. 😀
Ditto, if you’ve worked with numbers long and often enough, your instinct will give you some ideas.
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
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.
Yeah I thought this was quite an obvious way of doing it too?
thats what i did, the trickier part is proving that a perfect square will never end in these ways
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.
absolute mad lad
I agree@@Mk-rc7mj Absolute Mad lad. Wish I had enough mental fortitide to stick with such a daunting calculation.
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.
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
Leaving you on 69 likes coz that's dedication is nice
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.
Is this like elementary number theory or combinatorics or something? I wanna learn some before I take Discrete this fall
@@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?
@@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.
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.
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
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.
D is also divisible by 5 but not 25
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.
@@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.
Easiest way to eliminate E: a square cannot have an odd number of trailing zeroes.
In the last one I think you meant "the digits of E", but very nice
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)
I just eyeballed it and chose c.
@@rizzwan-42069 sad
Same thing I did. It will take just a glance if you know these.
I ruled out A B and E immediately and thought which one looked more like a square number was correct
Same here.
I ruled out A B and D immediately, then guessed C with the two left
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
I picked the same as well, thought I was wrong because I mistyped the 6th digit
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.
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
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.
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
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.
@ 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.
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.
@no, it’s k = n^2 * 10
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
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.
Take the last two digits and the middle 3 and your done
@@Ovoparity-jh6bt ???
I’m good at math, but this made my hair hurt
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
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.
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.
Mentioned in what?!?!?!
@@verygooddeal4436 It is bprp's Chinese channel.
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.
"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
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.
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.
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
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
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
Square Root of D is 26707.50896
Square Root of E is 31426.96295
Therefore, every number is a perfect square! Huh, silly Oxford!
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).
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
A 035 at the end will give 7 as last digit when divided by 5.. another way to rule out option d...
Amazing video! I will be trying to analyse big numbers like this now, very interesting!!
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
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
That's what I would have done if given this problem
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.
Your explanation for d) shouldn't take the 10's and 100's necessary equal to n. They are different in the problem
@@quantumgaming9180 n is only for the 10s. That is, (... + 10n + 5)*(... + 10n + 5) = ... + 100n^2 + 100n + 25
@@nathan87 ahhh I see now
(C) can be figured out using long division method for finding square roots.
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)
As n^2 = (50-n)^2 = (50+n)^2 (mod 100), the final two digits must occur in a square of a number below 26. This rules out a,b,d. If a square is divisible by 5^3 it is divisible by 5^4, which rules out e. Hence the answer is c.
The square root is 25,485. Took me a bit to reconstruct and execute the algorithm.
c)
Ez
Basically use properties of 3 and 9 multiples
Doing square roots without a calculator is similar to long division.
nuh uh
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
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
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.
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.
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.
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
For d) Since 713,291,035 is divisible by 5, for it to be a perfect square, it must be divisible by 25, which it is not.
For e) √(k•1000) can be a perfect square only if k is a multiple of 10. You can write 987,654,000 = k • 1000 where k is not a multiple of 5. But then 1000 = 8 • 5^3. So 987,654,000 = m • 5^3 where 5 does not divide m. But 5 is raised to an odd number not an even number.
I employed elimination:
1. Perfect squares *never* leave a remainder of 3 when divided by 4. This rules out A and D.
2. If a perfect square ends with 0, the number of trailing zeroes must be even. This rules out E.
3. If a perfect square is divisible by 3, it must be divisible by 9. 123,333,333 is divisible by 3 but not by 9, as seen in its digit sum 24 which is divisible by 3 but not by 9. This rules out B.
Hence the only possible perfect square is *C*
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)
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
I have just watched the video in 黑筆紅筆 2 hours ago, and I am now going to watch again.
Thanks, haha!
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
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.
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.
you mean last 3 digits. good fact to remember!
@@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!
It's funny for me how in school we are told not to guess the answer and Oxford comes with this exercise 😂
It's not guessing.
You receive the information that one of the five numbers is a perfect square. So if you can show for four of them that they cannot be a perfect square, you don't have to guess; you can be sure that the remaining number is the correct answer (unless someone made an error).
D and E: divisible by 5 and 100 but not their squuares, respectively. B: no square ends in 3. By vibes, we can judge that C is the correct answer between A and C
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).
A)1 below a square number (10000^2), the only 2 adjacent integers to be both square numbers are 0 and 1;
For the remaining numbers, I used the rule of "if a square number is divisible by prime p, it is also divisible by p^2":
B)Divisible by 3, but not 9, by sum of integers;
D)Divisible by 5, but not 25, by last 2 digits;
E)If divided by 100 (a quotient of 2 square numbers, if it is an integer, is also a square number), becomes divisible by 5, but not 25, by last 2 digits (40)
my reasoning for a, I can't think of any number that, when squared, becomes a number whose digits are all the same. B, same reasoning. D, squares of 5 always end in 25, so it cant be 35. And for E, it ends in 3 zeroes, when a number that ends in zero is squared, 1. It will always end in 0, 2. It will always have an even number of zeroes.
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
To find the square root of choice c, start by dividing it by 5^2=25 and 3^2=9. Those factors are present since c ends in 25 and has a digit sum of 45, a multiple of 9. Using long division, you get 649,485,225 = 3^2 * 5^2 * 2,886,601. The remaining factor is just a little less than 1700^2 = 2,890,000. Hmmm... How much less?
2,886,601 = 2,890,000 - 3399 = 2,890,000 - 3400 + 1 = 1700^2 - 2*1700 + 1 = (1700 - 1)^2 = 1699^2
So we have factored c into squares: 649,485,225 = 3^2 * 5^2 * 1699^2 = 25,485^2.
I would agree that getting rid of small, easily spotted, factors makes life a lot easier.
We’re calculating quadratic residue with this one🗣️🗣️🗣️
I do like that most of these methods are quick intuitive shortcuts, but they do feel a bit hand-wave-y. A more rigorous test would be that for any square number, every prime factor must have an even number of occurrences. (Quick proof: Suppose n is a positive integer and (a^j * b^k * c^l * ...) is its prime factorization. Then the prime factorization of n^2 is n^2 = n * n = (a^j * b^k * c^l * ...) * (a^j * b^k * c^l * ...) = a^2j * b^2k * c^2l * ... Therefore any square number has a prime factorization where each prime has an even number of occurrences.)
In order of simplicity:
e) 987,654,000 = 987,654 * 1000 = 987,654 * 2^3 * 5^3. Since 5 is not a factor of 987,654, we have exactly 3 occurrences of a prime factor 5. Therefore it is not a square number.
b) 123,333,333 = 41,111,111 * 3. By the digital sum test, 3 is not a factor of 41,111,111, and therefore we have exactly one prime factor 3. Not a square.
d) 713,291,035 = 142,658,207 * 5. Since 5 is not a factor of 142,658,207, we have exactly one prime factor 5. Not a square.
a) 99,999,999 = 3^2 * 11,111,111 = 3^2 * 11 * 1,010,101 = 3^2 * 11 * 101 * 10,001. 11 and 101 are both primes and neither is a factor of 10,001. (If you need to convince yourself of this, try substracting one of them from 10,001 and see if what remains has a factor. 10,001 - 11 = 9,990 = 3^2 * 1,110 = 3^2 * 2 * 5 * 111. 11 is not a factor of 111, and therefore is not a factor of the entire number.) Therefore it has prime factors which only occur once, and is not a square.
A) is right next to 10,000
Squared so it’s out, B) Ends in a 3 so it’s also out, E) has a odd number of zeros at the end so it’s out
That leaves C and D
D doesn’t work because you can only divide it by 5 once
And getting square root of C without a calculator is just a matter of brute forcing and making educated guesses and you would get it within a dozen or so tries
I found C pretty easily but calculating it took trial and error after getting the first few digits from 515^2 which led me to look ahead and see the 64 leading term and going from 8^2 to 25 as the leading term that got closest to 600,000,000 when squared without exceeding (25,000.) Then I just go closer by multiples of five before finding 25,485 as the answer.
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.
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.
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
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
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.
mod 4:
1^2 = 1
2^2 = 4 = 0
3^3 = 9 = 1
this rules out (a) and (d)
mod 5:
1^2 = 1
2^2 = 4
3^2 = 9 = 4
4^2 = 16 = 1
this rules out (b)
5^3 divides (e), and 5^4 does not divide (e)
a) Any perfect square that has 9 as the last number has an even number as the second to last number, so that's out (the ...3^2 cycle through 09, 69, 29, 89, 49, 09 ..., and the ...7^2 cycle through 49, 89, 29, 69, 09, 49,...). Also the 1 less than a perfect square thing, but my way was fun to prove to myself using the difference of squares once I realized it.
b) no square number ends in 3
d) all square number that end in 5 have 2 as the second to last number
e) All square numbers that end in 0 end in an even number of 0s.
The number before the 25 in a perfect square is the number before the 5 times (itself plus 1), so
6494852=x^2+x
2000
For the reason why the number before 9 is even, because the difference between x^2 and (x-1)^2 is 2x-1, it is either
00+9+3*10*some even number, i.e. 0+1+3+5=9 for 3^2 or 100+21+23+25=169 for 13^2
Or 25+4+2*some odd number, i.e. 25+11+13=49 for 7^2 or 225+31+33=289 for 17^2.
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
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
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.
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.
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.
For the actual answer, I'd just factor it so it comes out to something like (...x*y*z)^2 where each letter is a non-unique factor. It does get kind of difficult when you run into a 7-digit prime though.
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
Didn’t know that squares are all either 0 or 1 mod 4.
@@bobh6728 It is very easy to show, consider odd and even integers separately :)
@@nathan87 After I posted I looked it up. Just had never heard of that before. Even at my age, I can learn something new.
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.
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.
With respect to d, a square of a multiple of 5 must have 25 as a factor.
And with respect to E, a multiple of 3 must have 9 as a factor. (It doesn't.)
More than that: it must end in the two digits "25".
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
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
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
If you look at the final two numbers in each, the only one that works as a square is 25 (5 squared). Solved
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.
I also knew the answer, however, I have a hard time grasping fully why it couldn't be the one ending in 35 as well.
B, D, E saw instantly.
Took some time with A, didn't notice the method in the video so had to divide it by 9 and then by 11. The result wasn't dividable by 11 so I got my answer.
Striking out mismatching patterns, only c remains.
Matching paterns: even digit+"1", even digit+"4", even digit+"9", odd digit+"6", 25 and 00.
If you have "00", strike these digits and look what remains.
C is telegraphing its squareness. Not only is 25 a square number, so is 225 (15^2.)
richard
--
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
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
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
My method to look for the sqr. root would be:
1) Find if the number can be easily divided, so you count with smaller number in next steps. (int. factor. 3, 3, 5, 5, 2886601; counting with 2886601 in next steps)
2) Find number, that squared is close to number you're looking for. (It's bit less than 1700)
3) Find the exact value. Try only those numbers, which squared end with the same digit you're looking for. (1699)
4) You've got integer factorization of original number with all factors squared. Remove the square to get the answer you're looking for (3*5*1699).
For A, I used the rule for division by 11 (which is really awkward to recite, but it's a lot like the 3's one). Anyway 99,999,999 is divisible (99,999,999 = 11 x 9,090,909) but 9,090,909 isn't.
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.
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.
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.
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.
Another fun thing about squares is that the digital root of a square will only ever be 1, 4, 7, or 9
I would just assume the root is (100a + b), where b is some 2-digit number. Squaring it and taking mod 100, we have b^2 so we can eliminate a, b and d since their last 2 digits are not squares.
I guessed c because it had a 5 in the ones place so the squared number could have a 5 in its ones place since 5^2 = 25. You could make the same argument for d and you’d be right…
I've done large square roots on my abacus - schools don't teach it by hand anymore.
I asked my parents today about when they learned long-hand square roots in school. It was grade 4. Grade 4 in Canada used to teach this to kids.
I eliminated a,d because they are both equivalent to 3 mod 4, b because n^2 is never equal to 3 mod 10, and e because it is divisible by 5^3 (odd exponent). For 649 485 225 we can divide it by 25 to get 25 979 409, and can divide that by 9 to get 2 886 601. So we know that the square root of that will be between 1.6x10^3 = 2.56x10^6 and 1.7x10^3 = 2.89 - what's more, it'll be very close to 1700 (since 2890000-2886601 = 2889999-2886600 = 3399). What's more, for a square to finish in 01, the last two digits have to be 01, 49, 51, or 99 - and since all the numbers are clearly too small except 99, we get sqrt(2 886 601) = 1699 - and checking, 1699^2 = (1700-1)^2 = 2890000 - 3400 + 1 = 2886601. And the original number is a square of 1699*3*5 = (15*17)*100 - 15 = (16^2-1)*100 - 15 = 25500-15 = 25485
So I basically ruled out a, b and e due to the last 3 to 4 digits and I looked over at d which has 35 along with which isn’t a square and neither is 1035 so I went on to 25 which is a square of 5 and just to make sure I looked over to 225 which is the square of 15 so I basically just went with numbers that had squares on its end as someone who has just gone into 10th
I try:
e can't be square because of ending in an odd number of 0s -- a sqrt(10) appears
a can trivially be partially factorised to 3 * 3 * 11 * 101 * 10001, which clearly isn't divisible by 11, so no square
b is divisible by 3 an odd number of times, so no square
After partially dividing d by 5, it's clear that the result ends in a 7, which isn't divisible by 5, so no square.
answer by elimination: C
There's also modular arithmetic, which is the proper way to approach questions like this, but is also boring
C looks the most correct
Damn I did it. It literally stands out from the others due to it's end
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.
There are many methods to compute the value of the square. Herons method might be the oldest. Newton's methods is also great.
A - one less than (10⁴)², so no
B - equal to 6 mod 9, which is not a quadratic residue
d) divides 5 exactly once, so not a square
e) divides 5 exactly thrice, no