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]
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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)?
]]>