What is the smallest prime divisor of $5^{23} + 7^{17}$?
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+118667 What is the smallest prime divisor of \(5^{23} + 7^{17}\)
Powers of 5 are
5, 125, 625, ..
Powers of 7 are
7, 49, 343, 2401, 16807
7^1 ends in 7
7^2 ends in 9
7^3 ends in 3
7^4 ends in 1
7^5 ends in 7
and the pattern will then repeat.
7 9 3 1 7 9 3 1 7 9 3 1 7 9 3 1 7
So 5^23 ends in 5 and 7^17 ends in 7
so
\(5^{23} + 7^{17}\) ends in the last digit of 7+5 Which is 2
Any number ending in 2 is divisable by 2 so the smallest prime divisor of that expression is 2
+532 We know that odd x odd = odd. Here's a quick proof:
Say the odd number are a = 2x + 1 and b = 2y + 1. Then, we have
ab = (2x + 1)(2y + 1) = 2y(2x + 1) + 2x + 1 => 2(2xy + y) + 2x + 1 = even + odd = odd.
We know 5^23 = 5 * 5 * 5 * 5 * ..... * 5, with 23 5's. The first two fives pair up into an odd, and so on, so that product is odd. Repeat for 7^17 to get two odd numbers, and we know that odd + odd = even, so the answer is 2.
