N × U × (M + B + E + R) = 33
The only factors of 33 are 1,3,11,33
N × U must be one of the factors and M + B + E + R must be another.
N × U can only be 3,
for if it were 1, M + B + E + R would have to be 33 -- impossible!
and if it were 11, M + B + E + R would have to be 3 -- also impossible!
and if it were 33, M + B + E + R would have to be 1 -- also impossible!
So N and U are 1 and 3, and M + B + E + R = 11
So we must pick four digits from 0,2,4,5,6,7,8,9
that have sum 11.
The only ones that have sum 11 are the smallest four,
0 + 2 + 4 + 5 = 11
So the first way is
1 × 3 × (0 + 2 + 4 + 5) = 33
The 1 and 3 can be arranged in 2! = 2 ways
The 0,2,4, and 5 can be arranged in 4! = 24 ways
Answer: 2! × 4! = 2 × 24 = 48
Edwin
