We need to find all positive integers $N \le 100000$ such that $N \equiv 21 \text{ (mod 100}$ . We can always write any positive number $N = 100k + 10m + n$ , then we have:

$\begin{aligned} N^2 & \equiv (100k + 10m + n)^2 \text{ (mod 21)} \\ & \equiv (10m + n)^2 \text{ (mod 21)} & \small \blue{\text{Only the last 2 digits of }N\text{ affect}} \\ & \equiv 100m^2 + 20mn + n^2 \text{ (mod 21)} & \small \blue{\text{the last 2 digits of }N^2} \\ & \equiv 20mn + n^2 \text{ (mod 21)} \end{aligned}$

This implies that $20mn + n^2 \equiv 21 \text{ (mod 100)}$ . Since $20mn + n^2 \equiv n^2 \text{ (mod 10)}$ , this means that only the last digit $n$ affects the last digit of $N^2$ . Therefore, it can only be $n = 1$ or $n=9$ .

\When $\begin{cases} n = 1 & \implies 20m + 1 \equiv 21 \text{ (mod 100)} & \implies \begin{cases} m = 1 \\ m = 6 \end{cases} \\ n = 9 & \implies 180m + 81 \equiv 80m - 19 \equiv 21 \text{ (mod 100)} & \implies \begin{cases} m = 3 \\ m = 8 \end{cases} \end{cases}$

Therefore all $N$ which end with $11$ , $61$ , $39$ , and $89$ has $N^2$ end with $21$ . Since there are $4$ in $100$ , $100000$ has $\boxed {4000}$ .

if we write the last 2 digits of all squares we see that the digits 00 and 25 repeat after ten places and all other digits cycle periodically so numbers with last digit as 00 or 25 can be calculated by dividing n by 10 and rounding down that is the numbers of squares till n squared ending with 25 or 00 as their last two digits are the digits of n removing its last one digit for example for n=23454 the number of squares ending with last two digits as 25 or 00 is 2345. now for all other numbers there are just 22 different last two digits which are possible for squares the last being 96 obtained by 24.as they cycle periodically the number of digits having them can be calculated by subtracting 2 times the digits of n(excluding last)and them diving by 20 and rounding down then a 1 can be added to the numbers till the last one which in case of 23454 is 16 as its square has last digit as 16 and during the last cycle 16 and numbers preceding it will gain one more value.

Now for the actual question numbers with 25 or 00 as last digit is 10000(one zero less) so for other numbers it is (100000-2*10000)/20=4000 and as it ends with zero (a complete cycle)we do not need to add one and hence 4000 is the answer.

i think someone can generalise for base and for no of digits to be checked .is case of base 10 we have its factors as 5 whose square is 25 so it comes frequently and in this way someone can generalise this for any base.

feel free to be critical toward my approach i would like to hear everyone's opinions.

thanks for reading :)

Hi @Hamza Anushath I had deleted the post from my original account i.e @Siddharth Chakravarty so I couldn't write again.

Solution:

The last 2 digit of the square of a number depends on the last 2 digits of that number. So to get 21 as the last two digits we can only have 11, 39, 61 and 89 as the last 2 digits of the no. So the numbers would be

• 11, 111. 211. 311, 411, ...

• 39, 139, 239, 339, 439, ...

• 61, 161, 261, 361, 461, ...

• 89, 189, 289, 389, 489, ...

These above 4 sequences are A.P and they all have 1000 terms in between 100000. So the total no. of terms will be 4 x 1000 = 4000.

@Chew-Seong Cheong has already posted a proof for this. I did it using an another approach. I will soon upload it within a day as I'm trying to make some images or animation for the explanation why only 11, 39, 61 and 89 are possible, in short, I will provide a concise explanation.

Code

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j=0
for x in range(1,100000):
    if ((x**2)%100==21):
        j+=1
print (j)

Output

1

4000

@Mahdi Raza , @Vinayak Srivastava , @Páll Márton , @Siddharth Chakravarty , @Yajat Shamji

A question made in honour of you all