For every positive integer
n:
10^{n}+18*n1
is divisible by
27.
The above statement can be proven by more than one way.
Find at least 2 distinct methods.
One way:
10^n cycles through 10, 19, 1 ... mod 27, a cycle of length 3
18*n cycles through 18, 9, 0 ... mod 27, again a cycle of 3
The total, mod 27 is always 1, so when you subtract 1, the value is 0 mod 27, or in other words the number is divisible by 27.
Is this a different proof? ... proof by induction:
10^1 is congruent to 10 mod 27
18*1 is congruent to 18 mod 27
therefore, for n=1 10^n + 18*n  1 = 0 mod 27
10^(n+1)  10^n is 9 mod 27
18*(n+1)  18*n is 9 mod 27
therefore if 10^n + 18*n  1 = 0 mod 27 for n=k, then it is true for n = k+1.

Posted by Charlie
on 20171031 13:58:26 