Find the positive integer pair (m, n) such that m*n is the minimum and the last 98 digits in m^m*nⁿ are all zeroes.

Note: 99^th digit from the right in m^m*nⁿ is not zero.

The positive integer pair (m, n) is (98, 75) -- [and/or (75, 98)].

As m^m*nⁿ must trail with exactly 98 zeroes there must be at least 98 factors of one of the two numbers 2 and 5 and exactly 98 factors of the other.  The only way this can occur is if either m or n is 98 [98^98 = (2^98 * 7^196)], which has exactly 98 2s. Thus, the other number may be any number with 98 or more factors of 5 and no additional 2s -- or exactly 98 factors of 5.  Only multiples of 5 can have 5 as a factor, therefore the number must have greater than 98 factors of 5.  The smallest such positive number is 75 where 75^75 [5^150 * 3^75] has 150 factors of 5.

98⁹⁸*75⁷⁵ =
58850502064543772853645359113146360079940524370743
72805614890343159072210084737694472322605196104914
73752655063996301536290008346260850477235097238591
27768103676070524780090143856748233422088906083502
62407810760123538784682750701904296875000000000000
00000000000000000000000000000000000000000000000000
000000000000000000000000000000000000

Edited on October 30, 2007, 8:04 am

Posted by Dej Mar on 2007-10-28 11:57:14