Reaminder Problem 5

By Fermat's Little Theorem, $5^6 \equiv 1 \mod{7}$ , so $5^{99} \equiv 5^3 \equiv \fbox{6} \mod{7}$ .

$\begin{aligned} 5^{99} & \equiv (7-2)^{99} \text{ (mod 7)} \\ & \equiv -2^{99} \text{ (mod 7)} \\ & \equiv -8^{33} \text{ (mod 7)} \\ & \equiv -(7+1)^{33} \text{ (mod 7)} \\ & \equiv -1 \text{ (mod 7)} \\ & \equiv \boxed{6} \text{ (mod 7)} \end{aligned}$

From Fermat's Little Theorem, $\begin{array}{llr} &5^{99}&\mod7 \\=&5^{99\mod6}&\mod7 \\=&5^{(99-60)\mod6}&\mod7 \\=&5^{39\mod6}&\mod7 \\=&5^{(39-36)\mod6}&\mod7 \\=&5^3&\mod7 \\=&125&\mod7 \\=&(125-70)&\mod7 \\=&55&\mod7 \\=&(55-56)&\mod7 \\=&-1&\mod7 \\=&(-1+7)&\mod7 \\=&6 \end{array}$ It may be a bit long, but this is how I calculate it in my heart.

nice solution

Calculate the first few terms and look for a pattern:

>>> [(5**n)%7 for n in range(2,20)]

[4, 6, 2, 3, 1, 5, 4, 6, 2, 3, 1, 5, 4L, 6L, 2L, 3L, 1L, 5L]

Notice that it's mod 6.

99 mod 6 is 6.