What is the smallest prime divisor of $5^{23} + 7^{17}$?

Thank you for helping!!

Guest Mar 24, 2019

#1**+2 **

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

Melody Mar 24, 2019

#2**+3 **

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.

asdf335 Mar 24, 2019