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.

I am currently running a mathematica program that is test for values of the form

m=a*2^b*5^c

n=x*2^y*5^z

with a,x<=100 and b,c,y,z<=10 and I put in the current low for m*n of 10290 for it to try and beat.  So far it has spit out the following solution.

m=49*2=98   n=3*5^2=75

the requirements I am using to searh are the following

1)  a,x are not divisible by either 2 or 5

2) if p=min(b,c) and q=min(y,z) and r=min(b*m+y*n,c*m+z*n)

then p+q+r=98

the secound makes sure there are 98 factors of 10 in m^m*n^n and the first makes sure there are no extra factors of 2 or 5 and thus there would be exactly 98 zeros in m^m*n^n if both are satisfied.

I

Posted by Daniel on 2007-10-28 06:01:57