If p=1^3+3^3+5^3 +7^3 +9^3 +11^3 +13^3, what will be the

Solution 1:
To find the remainder when divide by 5, first we need to find the units digit.

\(1^3=1\) has units digit 1
\(3^3=27\) has units digit 7
\(5^3=125\) has units digit 5
\(7^3=343\) has units digit 3
\(9^3=729\) has units digit 9
\(11^3\) has units digit 1
\(13^3\) has the same units digit as \(3^3\), which has units digit 7

So the units digit of \(p\) is \(1+7+5+3+9+1+7=43\) or the units digit is 3.
So the remainder is 3 when divide \(p\) by 5.

Solution 2:
\(1^3 \equiv 1 \pmod{5}\)
\(3^3=(5-2)^3 \equiv -2^3 \pmod{5}\)
\(5^3 \equiv 0 \pmod{5}\)
\(7^3 =(5+2)^3 \equiv 2^3 \pmod{5}\)
\(9^3 =(5 \times 2-1)^3 \equiv -1^3 \pmod{5}\)
\(11^3 = (5 \times 2 +1)^3 \equiv 1^3 \pmod{5}\)
\(13^3 =(5 \times 3 -2)^3 \equiv -2^3 \pmod{5}\)

So \(p \equiv 1-2^3+0+2^3-1^3+1^3-2^3 \pmod{5}\)
\(\implies p \equiv 1-8=-7 \pmod{5} \implies p \equiv 3 \pmod{5}\)

The answer is D