And one more

Re: And one more

Bill Earl

This one was pretty easy to code up. Without the unique digit constraint, there are five solutions which satisfy the mathematical constraints. Four of them have one or more leading zeros.

Re: And one more

Larry Barrett

And of the five solutions, there are three with unique digits.

Limiting the trials

Alex Y

I took a slightly different tack. Not being a programmer, I was limited to a calculator or spreadsheet. But I only had to test a few cubes, from the cubes of 22 to 35, since 22 is the smallest integer greater than cube root of 10,234 (the smallest five-digit number with unique digits), and 35 is 9+8+7+6+5, the largest possible five-digit number with unique digits. Of these 14 cubes, only two met the arithmetic test, and only one of those had unique digits.

27³ = 19,683

Re: And one more

Bill Earl

Hmm. You must have found a different set of solutions than I did. I only get one solution with 5 unique digits.

Re: And one more

Larry Barrett

I also just used a spreadsheet. I found the same solution as Alex - 27 cubed = 19683

I also found 18 cubed = 05832, and 17 cubed = 04913. These seem like they work, but I have been wrong at least a couple of times before.

The other non-unique solutions I found are 8 cubed = 00512 and 1 cubed = 00001.

My mistake

Bill Earl

There was an error in my program that omitted a couple combinations. The complete set is:

00000

00001

00512

04913

05832

17576

19683

The highlighted ones meet the uniqueness test. Alex can decide if leading zeros are permitted.