Send More Money (Long Addition Puzzle)

BigBeachBanana

BigBeachBanana

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[math]\begin{array}{r}S\,E\,N\, D\\ +\,M\,O\,R\,E \\ \hline M\,O\,N\,E\,Y\\ \end{array}[/math]Solve for each distinct letter, where [imath]\text{ }M\neq 0[/imath]. The solution is unique. ?
Notice each subsequent column contains at least one letter from the previous.
Work from left to right
 
Last edited:
Since SEND + ONEY is the sum of two 4-digits numbers and it's at most 9,999+9,999=19,998 and M is not 0, then M must be equal to 1. I'll think about the rest.
 
I am assuming that the terms of the puzzle are that each roman letter is a distinct digit, with m not equal zero.

[math]1000s + 100e + 10n + d +\\ 1000m + 100\phi + 10r + e =\\ 10000m + 1000 \phi + 100 n + 10e + y.\\ 1000s + 100e + 10n + d < 9999\\ 1000m + 100\phi + 10r + e < 9999 \implies\\ 10000m + 1000\phi + 100n + 10e + y < 19998 \implies\\ m < 2 \implies m = 1 \ \because m \text { is a digit } \ne 0.[/math]
[math] 1000s + 100e + 10n + d +\\ 1000 + 100\phi + 10r + e =\\ 10000 + 1000 \phi + 100 n + 10e + y.\\ 1000s + 100e + 10n + d < 9999\\ 1000 + 100\phi + 10r + e < 2999 \implies\\ 10000 + 1000 \phi + 100n + 10e + y < 11998 \implies\\ \phi< 2 \implies\\ \phi = 0 \ \because \phi \text { is a digit } \ne m = 1.[/math]
[math]s < 9 \implies 1000s + 100e + 10n + d < 9000\\ 1000 + 10r + e < 1099 \implies\\ 10000 + 100n + 10e + y < 10099 \implies n \le 0,\\ \text {which is impossible } \because n \text { is a digit } \ne \phi = 0.\\ \therefore s = 9.[/math]
So far,

[math]\phi= 0,\ m = 1, \ s = 9, \text { and}\\ 9000 + 100e + 10n + d + 1000 + 10r + e =\\ 10000 + 100n + 10e + y \implies\\ 100e + 10(n + r) + (d + e) = 100n + 10e + y.[/math]
Moving on.

[math]d + e = y + 10 \alpha, \text { where } \alpha = 0 \text { or } 1,\\ \alpha + n + r = e + 10 \beta , \text { where } \beta = 0 \text { or } 1,\text { and }\\ \beta + e = n.\\ \beta = 0 \implies e = n, \text { which is impossible.}\\ \therefore \beta = 1 \implies n = e + 1 \implies\\ 100e + 10(n + r) + d + e = 100n + 10e + y \implies\\ 100e + 10(e + 1 + r) + d + e = 100(e + 1) + 10e + y \implies\\ 111e + 10 + 10r + d = 110e + 100 + y \implies 10r + d + e = 90 + y.\\ \text {But } d + e = y + 10 \alpha \implies\\ 10r + y + 10 \alpha = 90 + y \implies 10r + 10 \alpha = 90.\\ \alpha = 0 \implies 10r = 90 \implies r = 9 = s, \text { which is impossible.}\\ \therefore \alpha = 1 \implies 10r + 10 = 90 \implies r = 8 \text { and } d + e = y + 10. \therefore [/math]
Simplifying further, we have s = 9, r = 8, [imath]\phi[/imath] = 0, m = 1, and n = e + 1. So we are down to e, d, and y with many constraints so we can use brute force.
One constraint. is that d + e = y + 10 > 11. Another is 1 < d < 8. A third is 1 < e < 7.
If n < 6, then e < 5 so d + e < 12. Impossible.
If n = 7, then e = 6 and d < 6 so e + d < 12. Impossible.
If n = 6, e = 5, and d < 5 so e + d < 10. Impossible.

That leaves e = 5, n = 6, and d = 7 as the only possibility, which entails y = 2

Let's see whether s = 9, r = 8, [imath]\phi[/imath] = 0, m = 1, e = 5, n = 6, d = 7, and y = 2 works.

[math]1000s + 100e + 10n + d = 9567.\\ 1000m + 100\phi + 10r +e = 1085.\\ 10000m + 1000 \phi + 100n + 10e + y = 10652.\\ 9567 + 1085 = 10652. \ \checkmark[/math]
 
JeffM said:
I am assuming that the terms of the puzzle are that each roman letter is a distinct digit, with m not equal zero.

[math]1000s + 100e + 10n + d +\\ 1000m + 100\phi + 10r + e =\\ 10000m + 1000 \phi + 100 n + 10e + y.\\ 1000s + 100e + 10n + d < 9999\\ 1000m + 100\phi + 10r + e < 9999 \implies\\ 10000m + 1000\phi + 100n + 10e + y < 19998 \implies\\ m < 2 \implies m = 1 \ \because m \text { is a digit } \ne 0.[/math]
[math] 1000s + 100e + 10n + d +\\ 1000 + 100\phi + 10r + e =\\ 10000 + 1000 \phi + 100 n + 10e + y.\\ 1000s + 100e + 10n + d < 9999\\ 1000 + 100\phi + 10r + e < 2999 \implies\\ 10000 + 1000 \phi + 100n + 10e + y < 11998 \implies\\ \phi< 2 \implies\\ \phi = 0 \ \because \phi \text { is a digit } \ne m = 1.[/math]
[math]s < 9 \implies 1000s + 100e + 10n + d < 9000\\ 1000 + 10r + e < 1099 \implies\\ 10000 + 100n + 10e + y < 10099 \implies n \le 0,\\ \text {which is impossible } \because n \text { is a digit } \ne \phi = 0.\\ \therefore s = 9.[/math]
So far,

[math]\phi= 0,\ m = 1, \ s = 9, \text { and}\\ 9000 + 100e + 10n + d + 1000 + 10r + e =\\ 10000 + 100n + 10e + y \implies\\ 100e + 10(n + r) + (d + e) = 100n + 10e + y.[/math]
Moving on.

[math]d + e = y + 10 \alpha, \text { where } \alpha = 0 \text { or } 1,\\ \alpha + n + r = e + 10 \beta , \text { where } \beta = 0 \text { or } 1,\text { and }\\ \beta + e = n.\\ \beta = 0 \implies e = n, \text { which is impossible.}\\ \therefore \beta = 1 \implies n = e + 1 \implies\\ 100e + 10(n + r) + d + e = 100n + 10e + y \implies\\ 100e + 10(e + 1 + r) + d + e = 100(e + 1) + 10e + y \implies\\ 111e + 10 + 10r + d = 110e + 100 + y \implies 10r + d + e = 90 + y.\\ \text {But } d + e = y + 10 \alpha \implies\\ 10r + y + 10 \alpha = 90 + y \implies 10r + 10 \alpha = 90.\\ \alpha = 0 \implies 10r = 90 \implies r = 9 = s, \text { which is impossible.}\\ \therefore \alpha = 1 \implies 10r + 10 = 90 \implies r = 8 \text { and } d + e = y + 10. \therefore [/math]
Simplifying further, we have s = 9, r = 8, [imath]\phi[/imath] = 0, m = 1, and n = e + 1. So we are down to e, d, and y with many constraints so we can use brute force.
One constraint. is that d + e = y + 10 > 11. Another is 1 < d < 8. A third is 1 < e < 7.
If n < 6, then e < 5 so d + e < 12. Impossible.
If n = 7, then e = 6 and d < 6 so e + d < 12. Impossible.
If n = 6, e = 5, and d < 5 so e + d < 10. Impossible.

That leaves e = 5, n = 6, and d = 7 as the only possibility, which entails y = 2

Let's see whether s = 9, r = 8, [imath]\phi[/imath] = 0, m = 1, e = 5, n = 6, d = 7, and y = 2 works.

[math]1000s + 100e + 10n + d = 9567.\\ 1000m + 100\phi + 10r +e = 1085.\\ 10000m + 1000 \phi + 100n + 10e + y = 10652.\\ 9567 + 1085 = 10652. \ \checkmark[/math]
Click to expand...
That is the answer, one and only. Well done.?
 
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