For every positive integer
n:
10n+18*n-1
is divisible by
27.
The above statement can be proven by more than one way.
Find at least 2 distinct methods.
(In reply to
Puzzle Solution: Method 1 by K Sengupta)
10^1 = 10 (mod 27)
10^2 = 19 (mod 27)
10^3 = 1 (mod 27)
10^4 = 10 (mod 27)
......... and, so on.
Accordingly, 10^n (mod 27) cycles as: {10, 19,1, .....}
Again:
18*1 = 18 (mod 27)
18*2 = 9( mod 27)
18*3 = 0 (mod 27)
18*4 = 18 (mod 27)
....... and, so on.
Accordingly, 18*n (mod 27) cycles as: { 18,9,0,....}
Hence, 10^n+18*n-1 ( mod 27) cycles as {27,27, 0, ......}
or, 10^n+18*n-1 cycles as {0,0,0,.....} in mod 27 system.
Consequently, 10^n+18*n-1 is divisible by 27.
Puzzle Solution: Method 2
