Divide by $5$

Number Theory Level 2

If $n$ is a positive integer, then $n$ has $5$ possible remainders after division by five: $0, 1, 2, 3$ and $4$ .

If $n$ is a positive integer, then $n^2$ is also a positive integer.

$n$ is a positive integer. What is the number of possible remainders of $n^2$ after division by five?

2

5

1

4

25

3

2 solutions

Arjen Vreugdenhil
Dec 28, 2017

If $n = 5d + r$ , then $n^2 = 5(5d^2 + 2rd) + r^2,$ with $r = 0^2, 1^2, 2^2, 3^2, 4^2 = 0, 1, 4, 9, 16$ . Dividing these by five we get the remainders $0, 1, 4, 4, 1$ : $\boxed{\text{three}}$ different values.

Not satisfied with the solution

A Former Brilliant Member - 3 years, 5 months ago

Can you elaborate?

Arjen Vreugdenhil - 3 years, 5 months ago

An alternative approach: note that a perfect square ends in $0, 1, 4, 5, 6, 9$ . After division by 5, this results in the remainders $0,1, 4$ .

Arjen Vreugdenhil - 3 years, 5 months ago

Hey, zero is not a positive integer.

. . - 2 months, 1 week ago

No, but it is a possible remainder of dividing a positive integer by 5.

Arjen Vreugdenhil - 2 months, 1 week ago

. .
Mar 30, 2021

We only have to prove mod 5 in 5 numbers.

Firstly, $1 ^ { 2 } \mod 5 = 1$ .

Secondly, $2 ^ { 2 } \mod 5 = 4$ .

Thirdly, $3 ^ { 2 } \mod 5 = 4$ .

Fourthly, $4 ^ { 2 } \mod 5 = 1$ .

Fifthly, $5 ^ { 2 } \mod 5 = 0$ .

So, the possible numbers of remainder is $\boxed { 3 }$ .

Note: maybe you can think about zero, but zero is not a positive integer.

Divide by 5 5 5

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Divide by $5$