All digits of a positive number M are odd and no two are the same. All digits of the positive integer N are even and no two are the same. If N is a multiple of M, determine the largest possible value of M.

Question

All digits of a positive number M are odd and no two are the same.  All digits of the positive integer N are even and no two are the same.  If N is a multiple of M, determine the largest possible value of M.

asnaseer · Answer

largest M with those conditions must be 97531.
largest N would be 86420 (as 0 is considered even).
so largest NM = 86420 * 97531 = 8428629020
(I assume you wanted to know largest NM rather than largest M)

anonymous · Answer

I believe I want to know when M/N

asnaseer · Answer

in which case you want the largest M divided by the smallest N.
smallest N would be 2
so largest M/N = 97531 / 2 = 48765.5

anonymous · Answer

I think it has to go along the line of something like odd integers is 2M+1 and even intergers are 2N

asnaseer · Answer

sorry - I didn't read the "If N is a multiple of M" - let me try again :-)

anonymous · Answer

ok no problem!

asnaseer · Answer

ok, since N is a multiple of M, then M must be less than or equal to the (largest N) / (smallest multiple) = 86420 / 2 = 43210
so M < 43210, and using the conditions given:
largest M = 39751

asnaseer · Answer

correction, this doesn't satisfy the condition that N is a multiple of M with unique even digits.
M = 7531 is the largest M that yields
N = 8 * M = 60248