Square the Square #4th power

Number Theory Level 4

If $n$ is a perfect square, and ${ n }^{ 2 }\equiv k(mod\quad 10)$ what is the sum of all the possible values of $k$ and the number of values of $k$

Example: Suppose $k$ has $2$ values, $3$ and $4$ . Then the answer would be $2+3+4=9$

2 solutions

Alex Delhumeau
May 30, 2015

If $n$ is a perfect square, then its unit digit can be $0, 1, 4, 5, 6,$ or $9$ .

The units digit of $n^2$ can therefore be $0, 1, 5$ or $6$ for a total of $4$ possibilities.

$\Rightarrow 4+0+1+5+6=\boxed{16}$ .

Nelson Mandela
Feb 20, 2015

given, n is a perfect square.

n = l(mod 10) is nothing but the units digit.

A perfect square can only have 0,1,4,5,6 and 9 as their units digit.

The possible unit digit values of n^2 are 0(0^2), 1(1^2 and 9^2), 6(4^2 and 6^2) and 5(5^2).

So, total number of possibilities of k (n^2=k(mod 10)) is 4.

So answer is 4+0+5+6+1 = 16.

What about 4 and 9??

Ankit Kumar Jain - 6 years, 2 months ago

$CORRECT!!!!$

Vaibhav Prasad - 6 years, 3 months ago

i might be rude but you should have also mentioned about distinct digits.

abhigyan adarsh - 6 years, 1 month ago

I guess the ans is wrong. Any sqaure number modulo 10 gives remainders 0,1,4,5,6,9 ,Since there are 6 possibilities so the ans must be 31 not 16.

Debojyoti Biswas - 6 years, 1 month ago

The problem stated that $n$ is a perfect square, let's say that $n=k^2$ .

Hence, $n^2=k^4$ .

The possible unit digit of $k^4$ are $0,1,5,$ and $6$ .

Therefore, the answer to this problem is $0+1+5+6+4=16$ .

Ahmad Naufal Hakim - 6 years, 1 month ago