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

 

Thank you for helping!!

 Mar 24, 2019
 #1
avatar+109744 
+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  

 Mar 24, 2019
 #5
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+1

Thank you Melody, asdf335, and Chris!! 

Guest Mar 25, 2019
 #2
avatar+533 
+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.

 Mar 24, 2019
 #3
avatar+109744 
+3

asdf335    

That is a very good method !     cool

 

Using you method you can see straight off that  it will be 

odd+ odd which must be even 

So it has to be divisable by 2   

Melody  Mar 25, 2019
 #4
avatar+111396 
0

I like asdf's method, as well.....sometimes....we try to over-complicate things.......the reasoning is very simple and logical.....

 

 

cool cool cool

CPhill  Mar 25, 2019

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