# Verbal Arithmetic - Page 2

In column 4, the addition of S and M has produced a carry-over into column 5. The carry-over can only be a 1, since the letters represent single digit numbers. So for instance, if you add 8 and 9, you get 17, so you carry 1 into the next column. But you can never carry more than 1. So **M = 1**.

In order for there to be a carry from column 4 to column 5, S + M has to come to at least 9, so S is either 8 or 9. Therefore S + M is 9 or 10, and so O is 0 or 1. But we have established that M = 1, so therefore **O = 0**.

If there were a carry from column 3 to column 4, then E would be 9 and N would be 0. But O = 0, so there is no carry, and **S = 9**.

If there were no carry from column 2 to column 3, then given that E + O = N and also that we know that O = 0, this would mean that E = N. But that is impossible. So therefore there is a carry from column 2 to column 3, and N = E + 1.

If there were no carry from column 1 to column 2, then (N + R) mod 10 = E, and given that N = E + 1, this would mean that E + 1 + R = E mod 10, so therefore R would be 9. But we know that S = 9, so therefore there must be a carry from column 1 to 2, and **R = 8.**

To produce a carry from column 1 to column 2, D + E = 10 + Y. Since Y cannot be 0 or 1, D + E must be at least 12. As D is at most 7, then E is at least 5. Also, N is at most 7, and N = E + 1. So therefore E is 5 or 6.

If E were 6, then to make D + E at least 12, D would have to be 7. But N = E + 1, so N would also be 7, which is impossible. Therefore **E = 5 and N = 6.**

To make D + E at least 12, we must have **D = 7**, and so **Y = 2**.

A word of explanation about the use of mod 10. Modular arithmetic is sometimes referred to as clock arithmetic, although a clockface actually works in mod 12; so for example if it is 8 o'clock and you add 6 hours, you end up at 2 o'clock, so (8 + 6) mod 12 = 2. In the same way, if you add 7 and 8 you get 15. But if you are performing an addition, as we are here, you would carry the 1, and leave the 5. So therefore (7 + 8) mod 10 = 5.

This puzzle was originally set in the Strand magazine in 1924, by the well known English puzzler Henry Dudeney (1857 - 1930).

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