+0

+3
262
2
+132

What is the smallest prime divisor of (5^1999) + (6^1999) ?

a)2                    b) 5                     c)7                  d) 11                      e)13

Ok, this sounds like a very simple question. It is. But i am trying out for the maths olympiad tomorrow and I need to know how to do it without relying on the aid of calculating devices. Please help me ( and explain every step if you can).

Mar 13, 2018
edited by Jeffreymars16  Mar 13, 2018

#1
+18096
+1

Here is a START for you.....every power of 5 ends in 5  and every power of 6 ends in 6.....what happens when you add these together.......     will it be an odd number?  (not divisible by 2)   .....it probably won't be divisible by 5 will it?    Maybe 7?

What do YOU think???

By looking at the first few terms I see  11 is divisible in the 1st 3rd  5th 7th   9th  ....terms

while 2 3 5 7 are not.....I would guess 11.

Mar 13, 2018
edited by Guest  Mar 13, 2018
edited by ElectricPavlov  Mar 13, 2018
#2
0

Any sum of 5 + 6 both raised to an ODD power is divisible by the sum of 5 + 6 =11

Example: 5^3 + 6^3 =341 / 11 = 31......and so on. So, the smallest prime divisor of {5^1999 + 6^1999} is 11.

Mar 13, 2018