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

Mathematics Asked by user713999 on January 24, 2021

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?

6 Answers

You are correct that $n^2equiv7bmod100$ does not hold, but rather $n^2equiv70+k$.

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.

Correct answer by J. W. Tanner on January 24, 2021

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,

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

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

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

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

$text{*****}$


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 $6$).


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

Answered by CopyPasteIt on January 24, 2021

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 $76$.

Answered by fleablood on January 24, 2021

Hint

Look at the order of these solutions:

$quad 24^2 = 576$

$quad 74^2 = 5476$

$quad 26^2 = 676$

$quad 76^2 = 5776$

Answered by CopyPasteIt on January 24, 2021

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 $76$.

Answered by Barry Cipra on January 24, 2021

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 $25$.

Answered by N. S. on January 24, 2021

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