Question

How do you prove 6^7-6^6+6^5 is divisible by 31

Answer 1

Explanation:

I don't know if this is valid but

We can think of this problem like

(6^7-6^6+6^5)/31 = x

Where x is a whole number so 6^7-6^6+6^5 is divisible by 31

Multiply each side with 31 :

6^7-6^6+6^5 = 31x

We can turn 6^7 into 6^5 x 6^2

We can also turn 6^6 into 6^5 x 6

So the equation turns into this :

(6^5 x 6^2) - (6^5 x 6) + (6^5 x 1) = 31x

We can factor out the 6^5 out of the left side of the equation, which turns into :

(6^5) x (6^2 - 6 + 1) = 31x

Which is also :

(6^5) x (31) = 31x

Divide each side with 31

6^5 = x

X is obviously a whole number

Answer 2

the same digit. Please note that in order to check the last digit of a5
a
5
you just can ignore all digits but the last one. Thus, you can asume that a
a
has a single digit. You only have to check the fifth power of the numbers 0 to 9. It’s easy and fun, if I may add