By squaring 25,485 different numbers I have concluded that the square root of 649,485,225 is 25,485

I guessed C without even thinking about it, put me in Oxford

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.

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.

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.

I ruled out A B and E immediately and thought which one looked more like a square number was correct

Amazing video! I will be trying to analyse big numbers like this now, very interesting!!

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)

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

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.

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

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.

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

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.

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).

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

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.

All odd perfect squares are congruent to 1 mod 8 so you just need to check the first 3 digits mod 8 (e.g 999 mod 8) for the first four options.

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.

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

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.

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

Wow how exiting

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.

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

I have just watched the video in 黑筆紅筆 2 hours ago, and I am now going to watch again.

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 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.

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.

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.

Can we try linearization?

c) Ez Basically use properties of 3 and 9 multiples

Can you please make a video on vanishing factor

Doing square roots without a calculator is similar to long division.

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)

for eliminating option e, you can eliminate it because every square number than ends in 0 must end in an even number of zeros but 3 is not even

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

Saw the 25 and thought it was the best option.

C looks the most correct Damn I did it. It literally stands out from the others due to it's end

I looked at it for a couple seconds and knew it was C by instinct just by looking at the last 2 digits. There is no square number in existence that ends in 99, 33, 35, or 40 (e is a little wierd, I removed the last 2 0s to simplify it). That only leaves C, which ends in 25, a valid pair of digits for square numbers.

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

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.

(C) can be figured out using long division method for finding square roots.

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.

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)

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

Could you use the unique prime factorisation of numbers to solve this?

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)

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.

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.

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).

Another fun thing about squares is that the digital root of a square will only ever be 1, 4, 7, or 9

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.

With respect to d, a square of a multiple of 5 must have 25 as a factor.

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

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.

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

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.

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*

Please try the rest of the MAT questions from this year!!! The Fibonacci one was nice

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.

If you look at the final two numbers in each, the only one that works as a square is 25 (5 squared). Solved

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.

We’re calculating quadratic residue with this one🗣️🗣️🗣️

Another reason to reject b): it is divisible by 3 but not by 9 (3^2).

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

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 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…

Knowkng the answer is C, we know that 25000² is 625000000, so 25500² is 625000000 + 25000000 + 250000, or 650250000. Now, the option C is around 800000 less than that, so we divide 800000/(2*25000) and we get around 16. So the real square root is really close to 25500-16, or 25484. The nearest result is 25485, so that would be the square root

For options A and D, wasn't it just easier to say that n² is congruent to 0 or 1 (mod 4), and therefore they can be ruled out much faster?

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.

I've done large square roots on my abacus - schools don't teach it by hand anymore.

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

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 did this one in my head.

I looked at everything, noticed c was the only one that ended in a double digit square, and e didn’t look like the answer, so I picked c in less than 5 seconds.

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.

i eliminated a b and d just like you but since i didnt know how to eliminate e i just thought "c is definitely it e isnt in most of the mcqs so yeah"

i just tried to prime factor the numbers until i got a prime factor that appeared an odd amount of times bc i dont know all those rules. does anybody have the proof to why a square number can't end with a 3?

Looked at c saw it had 64 , 48, 225 , 25. Just felt right

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.

I used my FEELING and i actually got it correct , im surprised

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

It would have been much harder if they'd said "One _or none_ of these numbers is a perfect square." Positively proving (c) is a square is much harder than eliminating the others. With numbers ending in 5, not only does the tens digit have to be 2, but also the hundreds digit can only be 0, 2, or 6, and if it's 6 then the thousands digit can only be 0 or 5.

C is telegraphing its squareness. Not only is 25 a square number, so is 225 (15^2.) richard --

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.

Can’t believe I got this so quickly.

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 took this test back in October and got a 53%

just use long division, it is faster and takes less effort

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

I actually didnt know about the fact on which d option is eliminated Thanks for it ❤❤ And last option was eliminated due to odd no. Of zeros and first eliminated that

I guessed the correct one solely because it had a 25 at the end of it, so it the highest chance of being right. I also knew that it couldnt end in 3, but i couldnt explain why.

I figured that B is divisible by 3 but not by 9 so it cannot be. I used similar logic for D and E. For A I just guessed that 11 111 111 cannot be.

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.

(10n+5)^2 = 100n(n+1)+25

My first guess was 25,000^2 which is 625,000,000 so from there i'd just start guessing and use binary search. obviously it ends with a 5, too. Took me about 7 tries... not fun but no calculator. got up to 25505 then backed down by ten a few times

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

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.