20
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The well-known verbal arithmetic problem

    S E N D
+   M O R E
-----------
  M O N E Y

explains (or at least it did the first time I knew about it) the story of a student who send a telegram to his father, asking for more money, and asks to fill in each letter with a different digit in order to figure out how much the student was asking for. What's not so broadly known is the answer his father gave:

  S P E N D
-   L E S S
-----------
  M O N E Y

Even though, the father was not as good as the son with verbal arithmetics and by mistake made an impossible one.

Could you prove that the rest he proposed is impossible?

Of course, feel free to solve the first verbal arithmetic (which can be solved indeed), but the real question is about the impossibility to solve the second one.

Note: The first verbal arithmetic problem is only provided for flavour, and the same letter does not necessarily represent the same number between both equations.

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  • $\begingroup$ I've seen a different formulation of this one before. Good puzzle though. $\endgroup$ – Joe Z. May 27 '15 at 20:05
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    $\begingroup$ With the constraints stated here, the rest is possible, as 00000-0000=00000. You should explicitly state the constraints on the original problem, assuming they're the same for the new one. In particular, does each letter have to represent a different digit? $\endgroup$ – Jason C May 27 '15 at 20:06
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    $\begingroup$ @JasonC edited to clarify different letters are different numbers $\endgroup$ – Masclins May 27 '15 at 20:08
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    $\begingroup$ @JoeZ. The OP has added the constraint. Without the constraint, it's not impossible (e.g. wrt your answer, consider if N and S are both 0). $\endgroup$ – Jason C May 27 '15 at 20:08
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    $\begingroup$ I heard the response as SPEND-LESS=SON which also fails. $\endgroup$ – Ross Millikan May 27 '15 at 21:03
19
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If $E - E = N$ in the third column, then $N$ is either $0$ or $9$.

Suppose $N$ was $0$; then you'd have to borrow 1 in the fourth column to subtract $S$ from $0$, but $N$ can only be $0$ in the third column if you don't borrow anything.

Similarly, suppose $N$ was $9$; then you'd have to have borrowed 1 in the fourth column. But $9$ minus anything never borrows, so that case is impossible as well.

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    $\begingroup$ This logic works for any positive integer base, not just decimal numbers. N must either be 0 or (base -1), and won't work for the same reasons. $\endgroup$ – JAQFrost May 27 '15 at 21:49
2
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Here is a more non-technical:

Give each word a value of (+1) or (-1) for relevance to their meaning
Spend = -1
Send = +1
More = +1
Less = -1


Therefore [Send + more]: (+1) + (+1) = Money
and [Spend - Less]: (-1) - (-1) = Money
meaning that money must equal both 2, and 0. (Which is not possible)

The Dad should have said
|Spend + Less = Money|
|(-1) + (-1)| = |-2| = 2

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  • 1
    $\begingroup$ That's a rather easy way to solve it, provided both cryptograms were true at the same time. Even though, as the last note on the question stated, they are completely independent. $\endgroup$ – Masclins May 27 '15 at 20:17
  • $\begingroup$ @AlbertMasclans That is with dealing with letters, not words ;D $\endgroup$ – Mark N May 27 '15 at 20:18

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