Prove that 5^39 + 1 is divisible by 7

Answer 1

5^39 = 1.818989404 x 10^27 = 2.598556291 x 10^26

5^2 = 25.

5^3 = 125

5^4 = 625

5^5 = 3125

5^6 = 15625

5*7= 78125

5^8 = 390625

5^9 = 1953125

5^10 = 9765625

5^11 = 48828125

5^12= 244140625

5^13 = 122070325

5^39 = (5^13)^3 = (122070325)^3 = 1.49011645 x10^16 x 122070325 =

add 1 for 5^39 + 1

the number is divisible by 7 if you take the last digit, double it, subtract that from the remaining number and it is either 0 or divides evenly by 7 then the original number is divisible by 7

28, double 8 to get 16, 2-16= -14, double 4 to get 8 1-8 =-7 which is divisible evenly by 7

714 double 4 to get 8, 71-8 = 63 double 3 to get 6 6-6 = 0

351428 double 8 to get 16, 35142-16 = 35126 double 6 to get 12. 3512-12 = 3500. double 0

leaving 350 double 0 leaving 35 double 5 to get 10 3-10 =-7 which is evenly divisible by 7