last digit is a square and…

I've found some solutions for this questions but they were not impressive.

Question:

How many natural numbers are there in base $10$,whose last digit is perfect square,combination of last two digits is a perfect square,combination of last three digits is a perfect square,$\ldots$,combination of last $n$ digits is a perfect square?

For example $64$ is a number whose last digit is a perfect square and combination of last two digits is also a perfect square.

Kindly tell me how to approach this question.

• There are infinitely many such numbers. Here are some of them: $10^2,10^4,10^6,\dots$ – barak manos Dec 5 '16 at 10:35
• Try to write on a piece of paper some perfect squares, select those satisfying this property and look for a pattern. – Crostul Dec 5 '16 at 10:35
• I've done the same as suggested by crostul when I left the cases of number having only 0 & 1 as their digits. – Atul Mishra Dec 5 '16 at 10:37
• Your question starts with "How many natural numbers are there..." - well, there are infinitely many. What more do you want? – barak manos Dec 5 '16 at 10:38
• I want other than those numbers you have suggested, please,,,, – Atul Mishra Dec 5 '16 at 10:39

Answers are in the forms:-

(i)$4\times10^n$

(ii)$9\times10^n$

(iii)$10^n$

(iv)$49\times10^n$

(v)$64\times10^n$

(vi)$81\times10^n$

Where $n \in 0,2,4,6...$

• I beg your pardon? – Alex M. Dec 6 '16 at 15:05
• It's hard to understand what you mean (and you're wrong). – Henrik Dec 6 '16 at 15:50
• But I think these are the only possible solutions – Harsh Kumar Dec 6 '16 at 15:58
• What about $900$? – Henrik Dec 6 '16 at 15:58
• 900=9*10^2 @Henrik – Vidyanshu Mishra Dec 6 '16 at 16:06

Let us start without accepting trailing zeroes.

And let us then start by looking at last (least significant) digits.

As the last digit has to be a perfect square, it has to be $1$, $4$ or $9$.

Going to two digits, the above criterion, limits the number of candidates to $9\cdot 3 = 27$. Comparing those to a list of perfect sqaures, gives three numbers: $49$, $64$ and $81$.

If we go to three digits we again only have $27$ candidates, none of which are perfect squares.

So without trailing zeroes, there are six such numbers.

As $(ab)^2=a^2b^2$, it's clear that multiplying any two squares we get another square, and since multiplying by $10^n$ just adds $n$ zeroes to a number, we can multiply each of those six numbers by $10^{2n}$ for any $n\in\mathbb N$. Giving us all the solutions to the problem.