Send More Money (Long Addition Puzzle)

BigBeachBanana

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SEND+MOREMONEY\begin{array}{r}S\,E\,N\, D\\ +\,M\,O\,R\,E \\ \hline M\,O\,N\,E\,Y\\ \end{array}Solve for each distinct letter, where  M0\text{ }M\neq 0. The solution is unique. ?
Notice each subsequent column contains at least one letter from the previous.
Work from left to right
 
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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.

1000s+100e+10n+d+1000m+100ϕ+10r+e=10000m+1000ϕ+100n+10e+y.1000s+100e+10n+d<99991000m+100ϕ+10r+e<9999    10000m+1000ϕ+100n+10e+y<19998    m<2    m=1 m is a digit 0.1000s + 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.
1000s+100e+10n+d+1000+100ϕ+10r+e=10000+1000ϕ+100n+10e+y.1000s+100e+10n+d<99991000+100ϕ+10r+e<2999    10000+1000ϕ+100n+10e+y<11998    ϕ<2    ϕ=0 ϕ is a digit m=1. 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.
s<9    1000s+100e+10n+d<90001000+10r+e<1099    10000+100n+10e+y<10099    n0,which is impossible n is a digit ϕ=0.s=9.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.
So far,

ϕ=0, m=1, s=9, and9000+100e+10n+d+1000+10r+e=10000+100n+10e+y    100e+10(n+r)+(d+e)=100n+10e+y.\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.
Moving on.

d+e=y+10α, where α=0 or 1,α+n+r=e+10β, where β=0 or 1, and β+e=n.β=0    e=n, which is impossible.β=1    n=e+1    100e+10(n+r)+d+e=100n+10e+y    100e+10(e+1+r)+d+e=100(e+1)+10e+y    111e+10+10r+d=110e+100+y    10r+d+e=90+y.But d+e=y+10α    10r+y+10α=90+y    10r+10α=90.α=0    10r=90    r=9=s, which is impossible.α=1    10r+10=90    r=8 and d+e=y+10.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
Simplifying further, we have s = 9, r = 8, ϕ\phi = 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, ϕ\phi = 0, m = 1, e = 5, n = 6, d = 7, and y = 2 works.

1000s+100e+10n+d=9567.1000m+100ϕ+10r+e=1085.10000m+1000ϕ+100n+10e+y=10652.9567+1085=10652. 1000s + 100e + 10n + d = 9567.\\ 1000m + 100\phi + 10r +e = 1085.\\ 10000m + 1000 \phi + 100n + 10e + y = 10652.\\ 9567 + 1085 = 10652. \ \checkmark
 
I am assuming that the terms of the puzzle are that each roman letter is a distinct digit, with m not equal zero.

1000s+100e+10n+d+1000m+100ϕ+10r+e=10000m+1000ϕ+100n+10e+y.1000s+100e+10n+d<99991000m+100ϕ+10r+e<9999    10000m+1000ϕ+100n+10e+y<19998    m<2    m=1 m is a digit 0.1000s + 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.
1000s+100e+10n+d+1000+100ϕ+10r+e=10000+1000ϕ+100n+10e+y.1000s+100e+10n+d<99991000+100ϕ+10r+e<2999    10000+1000ϕ+100n+10e+y<11998    ϕ<2    ϕ=0 ϕ is a digit m=1. 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.
s<9    1000s+100e+10n+d<90001000+10r+e<1099    10000+100n+10e+y<10099    n0,which is impossible n is a digit ϕ=0.s=9.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.
So far,

ϕ=0, m=1, s=9, and9000+100e+10n+d+1000+10r+e=10000+100n+10e+y    100e+10(n+r)+(d+e)=100n+10e+y.\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.
Moving on.

d+e=y+10α, where α=0 or 1,α+n+r=e+10β, where β=0 or 1, and β+e=n.β=0    e=n, which is impossible.β=1    n=e+1    100e+10(n+r)+d+e=100n+10e+y    100e+10(e+1+r)+d+e=100(e+1)+10e+y    111e+10+10r+d=110e+100+y    10r+d+e=90+y.But d+e=y+10α    10r+y+10α=90+y    10r+10α=90.α=0    10r=90    r=9=s, which is impossible.α=1    10r+10=90    r=8 and d+e=y+10.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
Simplifying further, we have s = 9, r = 8, ϕ\phi = 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, ϕ\phi = 0, m = 1, e = 5, n = 6, d = 7, and y = 2 works.

1000s+100e+10n+d=9567.1000m+100ϕ+10r+e=1085.10000m+1000ϕ+100n+10e+y=10652.9567+1085=10652. 1000s + 100e + 10n + d = 9567.\\ 1000m + 100\phi + 10r +e = 1085.\\ 10000m + 1000 \phi + 100n + 10e + y = 10652.\\ 9567 + 1085 = 10652. \ \checkmark
That is the answer, one and only. Well done.?
 
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