Page 1 of 1
Digits
Posted: January 24th, 2023, 11:01 am
by cinelli
If we multiply 78 by 345 we get 26910 and all ten digits appear once and once only. This puzzle is to find another set of a x b = c where a has two digits, b three digits and c five digits so that all ten digits 0 to 9 appear precisely once. In addition b should be an exact multiple of a.
Cinelli
Re: Digits
Posted: January 24th, 2023, 12:52 pm
by UncleEbenezer
The constraint that most easily narrows the problem space is the exact multiple. Given b = a*d, we can rewrite c=a*a*d, and programatically test a small number of candidates within range and where a has no common factor with 10.
27 x 594 = 16038
Re: Digits
Posted: February 2nd, 2023, 8:38 pm
by cinelli
UncleEbenezer has the right answer. I did wonder whether this problem could be solved without computer aid and came up with the following.
If we consider the digital roots of a, b and c, (d(a), etc) I reckon that d(a) and d(b) must be either
1 and 4
3 and 6
7 and 7
9 and 9
in either order. This would save 33% of the work in trials for a but quite a bit more for b when you consider that they can’t share a digit. I think this is just about do-able.
Cinelli