# Show that 21(18^2x) + 36(7^3x) is divisible by 19 for all positive integers 'x'.

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Show that `21(18^(2x))+36(7^(3x))` is divisible by 19 for `x in NN` (x>0):

Factor out the common 3:

`3[7(18^(2x))+12(7^(3x))]`

Rewrite the exponentials using the power to a power rule (i.e. `(a^m)^n=a^(mn)` ):

`3[7(324^x)+12(343^x)]`

Note that 343-19=324. So `324^x=(343-19)^x` . We can use the binomial expansion on the binomial:(We will ignore the sign of the terms as this will not affect the divisibility, e.g. if 19 divides 57 it also divides -57)

`(343-19)^x=343^x+343^(x-1)19+343^(x-2)19^2+...+343*19^(x-1)+19^x` The important thing to note is that every term is a multiple of 19 except `343^x` , so we can write the right hand side as `343^x+19k` for some `k in ZZ`

Thus we have `3[7(324^x)+12(343^x)]=3[7(343^x+19k)+12(343^x)]`

`=3[7(343^x)+19(7k)+12(343^x)]`

`=3[19(343^x)+19(7k)]`

`=19[3(343^x)+3(7k)]`

which is divisible by 19.

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is there a way to solve this using modulo???

thx