Math CentralQuandaries & Queries


Question from ROBIN, a parent:



Hi Robin.

Most of the time in these questions, each letter represents a different digit (for example, if E is 2, then F, G, and H cannot be 2). This wasn't stated in the question, so we'll try to solve it without making this assumption.


You know that anything times four must be even, so if the last digit of the product is E, then E is 0, 2, 4, 6 or 8.

If E is the first digit of a four digit factor and multiplying it by 4 doesn't create a 5 digit product, then E must be a low number: Either 0 or 2.

If E is a zero, then 4 x H ends in a zero, so H must be 5 (H cannot be zero because E is already zero). So we have FG5 x 4 = 5GF0. But this is impossible, because 4 times a three digit number cannot be more than 4000. So E is not a zero. Thus E must be a 2.

result: E = 2.

We have: 2FGH x 4 = HGF2. So 4 x H ends in a 2. That means 4 x 3 or 4 x 8, making 12 or 32. So H is either 3 or 8.

But H is also in the thousands column of the product. Since we are multiplying a four-digit number by 1, the first digit of the product must be at least a 4, so H cannot be a 3. Thus H is 8.

result: H = 8.

We have 2FG8 x 4 = 8GF2 now. This means the carry into the tens column is a 3 (because 4 x 8 = 32). Thus, the result of 4G + 1 ends in the digit F. So F must be an odd number.

As well, since the 2 in the thousands of the factor becomes an 8 in the thousands of the product, that means there is no "carry" into the thousands. So that means that 4F + (carry into the hundreds) < 10. That means F is 0, 1 or 2. Combine this with the fact F is odd and so F must be 1.

result: F = 1.

We have 21G8 x 4 = 8G12. So 4G + 3 ends in a 1. This would make G either 2 or 7. So is 2128 x 4 = 8212? No. Therefore G must be a 7.

result: G = 7.

Check: 2178 x 4 = 8712. Yes.

Stephen La Rocque.

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