Fun with Numbers: The Pan-Prime-Digit Cube Hypothesis

[David Hayden]

(08-07-2017 10:50 AM)AlexFekken Wrote:  there are some obvious ways to cut down the number of required calculations quite a bit, as compared to a naive brute force attack. E.g. you could calculate the number of different values of n^3 mod 1000 (and then mod 10^6, then mod 10^9) that only contain the required digits. I would expect that that could narrow down the search quite a bit...

[ Edit: I mistakenly thought that both the number N and the cube N^3 had to be pan-prime-digital. In reality only N^3 must be. The description below reflects my misunderstanding. The basic approach can still be applied to the problem though.]

That's the approach I took last night with a C++ program and BigInt library. Multiplying two numbers with N least significant digits will result in the N least significant digits in the product, regardless of what the more significant digits are. So if you start with the pan-prime-digits and cube them. You find that for any number N ending in 2, N^3 will end in 8, so it can't be a pan-prime-digital cube.

Thus any pan-prime-digit cube must end in 3, 5, or 7.

Next you prepend 2, 3, 5, and 7 to these suffixes and test them. Here you find that any pan-prime-digit cube must end in 25, 33, 25, 37, 53, 55, 75 or 77. The program repeats the process until there are no more suffixes. I hoped that the number would eventually shrink. Boy was I wrong.

My program ran all night. When I stopped it, it was checking 10 million+ 35-digit suffixes and hadn't found a solution yet. I was running it on a slowish computer at work. I'll try running it on my laptop which has more horsepower.

Joe can you confirm that the answer you found is more than 35 digits?