Bunuel wrote:

If p is a positive integer, and r is the units digit of \(3^{4p}×5^{5p}×6^{6p}×7^{7p}\), then r + 2 is divisible by which of the following?

A. 2

B. 3

C. 5

D. 7

E. 9

Without calculating the units digits of each value in the expression, we see that since 5^5p has a prime factor of 5 and 6^6p has a prime factor of 2, we know that r is 0, and thus r + 2 = 2, so r + 2 is divisible by 2.

Alternate Solution:

Let’s focus just on 5^5p and 6^6p. Note that the units digit of 5^5p is 5, no matter what the value of p, because 5^1 = 5, 5^2 = 25, and 5^3 = 125, and this pattern of 5 as the units digit of any power of 5 continues. Similarly, consider powers of 6: 6^1 = 6, 6^2 = 36, and 6^3 = 216, and we see that 6 raised to any power has a 6 as the units digit in the answer. The product 5 x 6 is 30, and we see that the units digit of the product 5^5p x 6^6p is 0.

Thus, no matter what the other numbers are in the product, we see that the long product 3^4p x 5^5p x 6^6p x 7^7p will end in 0, because 5^5p x 6^6p = 0. Thus, r = 0.

We can also see, then, that r + 2 = 2, which is divisible by 2.

Answer: A

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