A Remainder Problem

When we take any even number to the power of 5, we'll be multiplying the number by 2 at least twice, which means that it will leave a remainder of 0, modulo 4.

Thus the expression becomes $1^{5} + 3^{5} + 5^{5} + ... + 99^{5}$

Now, $1^{5} \equiv 1 (mod 4)$ and $3^{5} \equiv 3 (mod 4)$ .

$5^{5} \equiv (4 + 1)^{5} \equiv 1^{5} \equiv 1 (mod 4)$

$7^{5} \equiv (4 + 3)^{5} \equiv 3^{5} \equiv 3 (mod 4)$

$9^{5} \equiv (8 + 1)^{5} \equiv 1^{5} \equiv 1 (mod 4)$

$11^{5} \equiv (8 + 3)^{5} \equiv 3^{5} \equiv 3 (mod 4)$

...

$(4n+1)^{5} \equiv 1^{5} \equiv 1 (mod 4)$

$(4n+3)^{5} \equiv 3^{5} \equiv 3 (mod 4)$

Therefore, the expression becomes

$25 \times 1 + 25 \times 3 \equiv 25 \times 4 \equiv \boxed {0} (mod 4)$

Note than by Euclid's Division Lemma, any odd number can be expressed in the form of $(4n+1)$ or $(4n+3)$ . Thus these are the only possible forms of the powers of the odd numbers left in the sequence, except for $1^{5}$ and $3^{5}$

Could someone tell me how to create the space between the last number in the line and the $(mod 4)$ ?

consider, $1^{5}+2^{5}+3^{5}+.................+100^{5} \equiv x \pmod{4}$ $=>1+0+243+0+0+0+...........+0 \equiv x \pmod{4}$ $=>remainder=\frac{244}{4}$ therefore the remainder is 0.