Hi all,
Once again I'm pretty stumped at this one.
3^3 = 27
3^4 = 81
3^5 =243
3^6 = 729
3^7 = 2187
Now these are the same practically. An even second to last digit and a seven as the last digit. So the pattern will loop. Just proving it seems a right pain! Any help at all would be appreciated.
Base case n=3 then 3^3=27 and so it is true that the second least significant digit is even.
Suppose it true for some , that is the second least significant digit is even.
Now is odd, and not divisible by , so the least significant digit is one of:
Now the carry when these are multiplied by is even (either or ), and
times the second least significant digit of is even. Even plus an even is
even so the second least significant digit of is even, which
completes the induction step.
So we have proven that for all the second least significant digit is even.
RonL
Hello, Jimbobobo!
I have a slight variation of RonL's solution.
We note that ends in: 1, 3, 7, or 9.Prove by induction that in the decimal form of , the tens-digit is even.
And ends in: 03, 09, 21, or 27.
. . Hence, there is a "carry" of either 0 or 2.
Verify . . . true!
Assume has an even tens-digit.
. . This means that: . is of the form: .
. . Hence: .
Multiply by 3: .
. .
Therefore, has an even tens-digit.
. . The inductive proof is complete.
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I had this wild idea of using binary, but I got nowhere.
Consider: .
I thought that . would lead to a neat solution,
. . but I didn't find one.
Anyone care to give it a try?