Prove that 2(6^2n)-5(6^n)-2 is divisible by 10 for all values of n.

At n=0 that looks like 2(6^0)-5(6^0)-2 = -5, a counterexample.

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It works for n=1 and n=2 so let's try to prove it for natural n>0. 

I've been warned off using "mod" in answers so we'll kind of handwave around it.

We substitute n=m+1 so m is a non-negative integer, m >= 0.

The term times 30 will always be a multiple of 10, so we need to check the rest of it is as well.  The factor of 72=70+2 and the 70 part will always give a multiple of 10.   

Also and that will have the same remainder as when divided by 10.  In fact every power of 6 except the zeroth ends in 6, because 6 times 6 is 36.   

So we must show



is a multiple of 10 which of course it is.  We also need to check m=0, 2(1)-2=0, ok as well. That ends the proof.

Answer:

13.46

step-by-step explanation:

16.80 x .20 = 13.46