Solution

Since the unit digit of Y + N + N is Y, N = 0 or 5. No matter it is 0 or 5, the carry from the unit digit can only be 1 or 0.

Since the unit digit of T + E + E is T and E + E must be even, there is no carry from the unit digit to the tens digit (otherwise the T in ‘SIXTY’ and that in ‘FORTY’ will have different parity, and this is not possible).

Since there is no carry from the unit digit to the tens digit, N = 0.

Also as , so there is carry from the thousands digit to the ten thousands. Thus I = 0 or 1. However as N = 0, I = 1. Thus O = 9 and there is a carry of two from the hundreds digit to the thousands. Now the numbers 2, 3, 4, 6, 7 and 8 are not used. Since there is a carry of two from the hundreds digit to the thousands, T = 6, 7 or 8.

(1)    If T = 6, then R = 7 or 8.

        (a)    If R = 7, since there is a carry from the tens digit to the hundreds, R + T + T + 1 = 7 + 6 + 6 + 1 = 20, thus X = 0. However as N = 0, this is impossible.

(b)    If R = 8, we have X = 1 (similar to above). But since I = 1, this is also impossible.

Therefore .

(2)    If T = 7, then R = 6 or 8.

(a)    If R = 6, we have X = 1. But as I = 1, this is impossible.

(b)    If R = 8, we have X = 3. Now we have 2, 4 and 6 left. But since there is a carry of 1 from the thousands digit to the ten thousands, F and S are two consecutive integers. Thus R cannot be 8.

(3)    If T = 8, then R = 6 or 7.

(a)    If R = 6, we have X = 3. Similar to above, there are no consecutive integers left, this is impossible.

(b)    If R = 7, we have X = 4. Now we have 2, 3 and 6 left. Thus F = 2, S = 3 and Y = 6.

The answer is as below: