Question

Which one of the following numbers will completely divide \(4^{61} + 4^{62} + 4^{63} + 4^{64}\)?

• A. 3
• B. 10
• C. 11
• D. 13

Hint:

Factor out 4^61 to get 4^61 x 85, then check which option divides 2^122 x 5 x 17.

Answer 1

The Correct Option is B

A quicker route uses only the last digit (units digit) of each term, since divisibility by 10 depends only on whether the number ends in 0.

1. Find the units digit pattern of powers of 4: $4^1$ ends in 4, $4^2$ ends in 6, $4^3$ ends in 4, $4^4$ ends in 6. So odd powers of 4 end in 4, and even powers of 4 end in 6.
2. Since 61 is odd, $4^{61}$ ends in 4. Since 62 is even, $4^{62}$ ends in 6. Since 63 is odd, $4^{63}$ ends in 4. Since 64 is even, $4^{64}$ ends in 6.
3. Add the units digits: $4 + 6 + 4 + 6 = 20$, which ends in 0.
4. A sum ending in 0 is always divisible by 10 (and by 2 and 5, but not necessarily by other primes like 3, 11 or 13).
5. To rule out 3, 11 and 13 rigorously, note $4^{61}+4^{62}+4^{63}+4^{64} = 4^{61}(1+4+16+64) = 4^{61}\times 85 = 2^{122}\times 5\times 17$, which has no factor of 3, 11 or 13.
6. Therefore the number is divisible by 10 but not by 3, 11 or 13.

\[\boxed{10}\]