SOLUTION: Prove That: 5^31–5^29 is divisible by 100. Format: x^y*100

Algebra.Com
Question 1067632: Prove That:
5^31–5^29 is divisible by 100.
Format: x^y*100

Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Prove That:
5^31–5^29 is divisible by 100.
----------
5^31–5^29 = 5^29*(25 - 1)
= 24*25*5^27
= 600*5^27
= 6*5^27*100

RELATED QUESTIONS

Prove that n2 − 2 is not divisible by 5 (answered by ikleyn)
If a number is selected at random from 61 to 100. Find the probability that it is... (answered by Fombitz)
i am a multiple of 5 between 100 and 200 that is divisible by 3,4, and, 6 who am... (answered by Studenttt)
30x + 31( 5 /5-x)... (answered by Ryan O'Hara)
Prove that x^2 -x ,is divisible by 2 (answered by KMST,rothauserc)
Prove that n^5 − n is divisible by 5 for any natural... (answered by richard1234)
Prove that n^8 − n^4 is divisible by 5 for any natural... (answered by robertb,ikleyn)
prove that {{{16^4-2^13-4^5}}} is divisible by... (answered by ikleyn,MathTherapy)
Prove that... (answered by Alan3354)