Solution Since the carry from the thousands digit to the ten thousands is at most 1, we have E = 1. Since F is at most 9, and the carry from the hundreds to the thousands is at most 2, the I in EIGHT is either 1 or 0. But E = 1, so I must be 0. Now the I in FIVE is 0. So the carry from the hundreds to the thousands is at most 1. It follows that F = 9. Consider the unit digit. Since E = 1, we have O + 2 = T. Since T cannot be 0 or 1, O cannot be 8 or 9. Consider the hundreds digit. Since the carry to the thousands is 1, T + O is at least 10. There are thus three possibilities: (1) O = 4, T = 6; (2) O = 5, T = 7; (3) O = 6, T = 8. (1)       O = 4，T = 6。 We have T + O = 10. So G is the carry from the tens digit. Now the remaining digits are 2, 3, 5, 7, 8. Since G is not equal to 0 or 1, so G must be 2. Thus V, W, N must be 5, 7 and 8 in order that their sum being at least 20. Note that in the unit digit, there is no carry. Thus H is the unit digit of 5 + 7 + 8, which is 0. But I = 0, so this is not allowed. (2)       O = 5，T = 7。 The remaining undetermined digits are 2, 3, 4, 6, 8. In the hundreds digit, I + T + O = 12. The carry from the tens digit is at most 1. So G is either 2 or 3. If G = 2, the tens digit has to carry of the hundreds. But the smallest three digits, 3, 4 and 6, have a sum exceeding 10. This is thus not possible. If G = 3, then we consider the value of H. If H = 2, V, W, N are 4, 6, 8. But the unit digit of their sum is not 2, a contradiction. Similarly, we can show that H cannot be 4, 6 or 8. Thus G is not 3. (3)       O = 6，T = 8。 Using a method similar to that in (2), we get G = 5, H = 3. Thus V + W + N has the same unit digit as 7 + 4 + 2, i.e. 3. By checking, we get the following 6 solutions (by permuting the values of V, W and N). Answers: [<< Previous] Example 3 [Next >>] Example 4

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