To solve the problem of finding the remainder when \(5^{20}\) is divided by \(7\), we can use Fermat's Little Theorem. This theorem states that if \(p\) is a prime number and \(a\) is any integer not divisible by \(p\), then:
\(a^{p-1} \equiv 1 \pmod{p}\)
In this problem, \(a = 5\) and \(p = 7\). According to Fermat's Little Theorem:
\(5^{7-1} = 5^6 \equiv 1 \pmod{7}\)
So, \(5^6\) leaves a remainder of 1 when divided by 7. Next, we need to express \(5^{20}\) in terms of \(5^6\).
Notice that:
\(5^{20} = (5^6)^3 \times 5^2\)
We can calculate each term modulo 7:
\(5^2 = 25\) and dividing 25 by 7 leaves a remainder of 4, thus:
\(5^2 \equiv 4 \pmod{7}\)
Since \((5^6)^3 \equiv 1 \pmod{7}\), we have:
\(5^{20} \equiv 1 \times 4 \equiv 4 \pmod{7}\)
However, there is an error in calculating \((5^6)^3\\) operation earlier, let us correct:
From Fermat's Theorem, \(5^6 \equiv 1 \pmod{7}\), so
\((5^6)^3 \equiv 1^3 \equiv 1 \pmod{7}\)
Thus, recalculating correctly:
Finally, multiplying by remaining term modulo 5 gives \(5^{20} \equiv 1 \times 1 \equiv 1 \pmod{7}\)
Therefore, the remainder when \(5^{20}\) is divided by \(7\) is 1.