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

Question

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

So $n^2 \equiv 7 \pmod{100}$? If this is the case then this can be written as $n^2 = 100k +7$, where $k \in \Bbb Z.$

Here one can see that no matter what the choice of $k$, the units digit will be $7$. Thus $n^2 \equiv 7 \pmod{10}.$ However this was wrong. The correct answer is $\textbf{6}.$

What am I doing wrong here? It seems that $n^2 \equiv 7 \pmod{100}$ doesn't hold. If the tens digit is $7$ should I have that $n^2 \equiv 7k \pmod{100}$, where $k$ represents the unit digit of $70$ and not a multiplication?

Answer 1

You are correct that $n^2\equiv7\bmod100$ does not hold, but rather $n^2\equiv70+k$.

To be a square, the last two digits have to have remainder $0$ or $1$ when divided by $4$

and remainder $0, 1, $ or $4$ when divided by $5$.

Look at the numbers from $70$ to $79$, and figure out which one satisfies those

to figure out what the last digit of $n^2$ must be.

Answer 2

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 $$n^2 \equiv 2+ k \pmod{4}\\ n^2 \equiv k-5\pmod{25} $$

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 $25$.

Answer 3

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 $76$.

Answer 4

Hint

Look at the order of these solutions:

$\quad 24^2 = 576$

$\quad 74^2 = 5476$

$\quad 26^2 = 676$

$\quad 76^2 = 5776$

Answer 5

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 $n^2 \equiv 70 + x\pmod {100}$.

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 $8k +1\equiv 7$ 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 $76$.

Answer 6

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
$\; 34^2 = 1156$
$\; 36^2 = 1296$
$\; 44^2 = 1936$
$\; 46^2 = 2116$