1) Do some raising of a power and play spot the pattern.

2) Repeated squaring.

3) Use the binomial theorem.

1) Did not yield anything I could do. Which of 2 and 3 would you like?

]]>How do you do it?

[This problem is not meant for a computer]

]]>641

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Compute the last 3 digits of 171^172.]]>

I couldn't find any others either.

]]>bobbym wrote:You mean of the type

3^(xxxxxx)?

Nearly. It is y^(xxxxxx). x and y are single-digit integers >0, and y may = x.

So the test is this:

For a=y^(xxxxxx) and b=y(x+x+x+x+x+x), the middle digit for Length[a]=odd (or the middle two digits for Length[a]=even) = b.So far, after not looking any further than my example in post #14, all I've found is just that one solution.

There are no such numbers besides x=5 and y=3.

]]>Okay, let me know if you find one more.

]]>It was only something completely frivolous where the numbers just happened to fall into place, but now I've set it up like this I might see if there are other solutions...if only to exercise my M.

]]>You mean of the type

3^(xxxxxx)?

Nearly. It is y^(xxxxxx). x and y are single-digit integers >0, and y may = x.

So the test is this:

For a=y^(xxxxxx) and b=y(x+x+x+x+x+x), the middle digit for Length[a]=odd (or the middle two digits for Length[a]=even) = b.

So far, after not looking any further than my example in post #14, all I've found is just that one solution.

]]>3^(xxxxxx)?

]]>