



 
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. Begin: 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.
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 fourdigit 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.
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.
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.
Check: 2178 x 4 = 8712. Yes. Cheers,  


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