Let $n$ be an integer. If the tens digit of $n^2$ is 7, what is the units digit of $n^2$?

When preparing for contest math questions it is helpful to have some theoretical shortcuts memorized. For these type of questions the following two facts can be applied.

Proposition 1: If $s$ is any integer then $s^2 equiv (s+50)^2 pmod{100}$.

Proposition 2: The following four modulo statements are true,

$; 1^2 equiv 9^2 equiv 11^2 equiv 19^2 equiv 21^2 equiv 29^2 equiv 31^2 equiv 39^2 equiv 41^2 equiv 49^2 pmod{20}$

$; 2^2 equiv 8^2 equiv 12^2 equiv 18^2 equiv 22^2 equiv 28^2 equiv 32^2 equiv 38^2 equiv 42^2 equiv 48^2 pmod{20}$

$; 3^2 equiv 7^2 equiv 13^2 equiv 17^2 equiv 23^2 equiv 27^2 equiv 33^2 equiv 37^2 equiv 43^2 equiv 47^2 pmod{20}$

$; 4^2 equiv 6^2 equiv 14^2 equiv 16^2 equiv 24^2 equiv 26^2 equiv 34^2 equiv 36^2 equiv 44^2 equiv 46^2 pmod{20}$

$text{*****}$

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It is immediate that if $n equiv 0 pmod5$ then the tens digit of $n^2$ can't be equal to $7$.

Since $1^2 equiv 1 pmod{20}$ and $2^2 equiv 4 pmod{20}$ and $3^2 equiv 9 pmod{20}$ we know that a solution satisfies

$; n in {4, 6, 14, 16, 24, 26, 34, 36, 44, 46 }$

and that the units digit of $n^2$ must be equal to $6$
(if $m equiv 4^2 pmod{20}$ then the units digit of $m$ is $6$).

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To find the actual solutions less than $50$ you calculate,

$; 4^2 = 16$
$; 6^2 = 36$
$; 14^2 = 196$
$; 16^2 = 256$
$; 24^2 = 576$ - OK
$; 26^2 = 676$ - OK
$; 34^2 = 1156$
$; 36^2 = 1296$
$; 44^2 = 1936$
$; 46^2 = 2116$

The tens digit is $7$, not the units and you want to find the unit digit.

So if the unit digit is $x$ then the number ends with $7x$ and $pmod{100}$ what you are trying to say is $n^2 equiv 70 + xpmod {100}$.

The way I would do this is let $n= 10k + a$ where $a,k$ are single digits. And the hundreds place don't affect the last two digits we might as well assume $n$ has only two digits.

$n^2 = 100k^2 + 20ak + a^2 = 100m + 70 + x$.

As $7$ is odd but $2ak$ is even so $a^2$ must be two digits and we carried an odd digit. if $a = 0,1,2,3,4,5,6,7,8,9$ then $a^2 = 0,1,4,9,14,25,36,49,64,81$.. So $a = 4$ or $6$.

So if $a = 4$ and $n=10k +4$ then $n^2 = 100k^2 + 80k+ 16$ and if $a=6$ and $n=10k+6$ then $n62 = 100k^2 + 120k + 36$. In either event $x = 6$.

We can have $8k +1equiv 7$ and $k = 2,7$ or we can have $2k+3 equiv 7$ and $k=2,7$. Notice $24,26,74,76$ when squared all end with $76$.

Hint

Look at the order of these solutions:

$quad 24^2 = 576$

$quad 74^2 = 5476$

$quad 26^2 = 676$

$quad 76^2 = 5776$

Any number that is a square mod $100$ is necessarily a square both mod $4$ and mod $5$, which is to say $0$ or $1$ mod $4$ and $0$, $1$, or $4$ mod $5$. The only number in the $70$s that satisfies both criteria is $76$.

Hint You are looking for adigit $k$ such that $$n^2 equiv 70+ k pmod{100}$$

By the Chinese Remainder Theorem this is equivalent to $$n^2 equiv 2+ k pmod{4}\ n^2 equiv k-5pmod{25} $$

The quadratic residues modulo $4$ are $0,1$, therefore $k in { 2,3, 6,7 }$. You have now to figure for which of those $k-5$ is a quadratic residue modulo $25$.