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Prove by induction that for every n >= 1 we have 19^n - 13 is a multiple of 6.

User Sdwdaw
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Step-by-step explanation:

For n=1, 19^n -13 = 19 -13 = 6, a multiple of 6.

For n = n+1, ...

19^(n+1) -13 = 19(19^n -13) +18·13

the factor (19^n -13) is divisible by 6 (for n=1), and ...

the factor 18·13 = 6·39 is divisible by 6.

Hence 19^n -13 is divisible by 6 for all n ≥ 1.

_____

Another way to look at this is to do modulo 6 arithmetic on the expression:

19^n -13 mod 6 = (19 mod 6)^n - (13 mod 6) = 1^n - 1 = 0

If the expression mod 6 is 0, then the expression is divisible by 6.

User Himanshu Tanwar
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