What is the product of a one-digit number and a two-digit number using 8 4 1 7 3?

Answer 1

This is an interesting problem.

First of all, 1 can definitely not be the 1-digit number, since the other 4 digits would then be grouped into 2 pairs.

Next, if 3 is the 1-digit number, we need to arrange 1, 4, 7, and 8 into two 2-digit numbers such that 3 times one of them equals the other. The first number (the smaller one) must have a 1 in the tens' place, since any other digit there would give a 3-digit result, not a 2-digit one. We have three cases: 3*14 = 42, 3*17 = 51, and 3*18 = 54. Neither is a solution.

If 4 is the 1-digit number, we need to arrange 1, 3, 7, and 8 into two 2-digit numbers such that 4 times one of them equals the other. Again, the first number must have a 1 in the tens' place (by the same reasoning as above), so there are three more cases: 4*13 = 52, 4*17 = 68, and 4*18 = 72. The last one is close, but still doesn't match the exact digits that were given.

Note that 8 cannot be the 1-digit number, since even with a 1 in the tens' place of the lesser 2-digit factor, the result would equal at least 8*13 = 104.

So we now need to check the three 7*1x products, but neither of those is a solution...

So, in conclusion, the original problem is either not properly explained, or contains a typo (is there a 2 instead of the 3?), or it simply does not have a solution.

New interpretation

If the digits are allowed to repeat, then the only answer is 4 * 37 = 148.