53 of 100: A Gram From Mars

From ones place $S+R+A$ is divisible by $10$ and since $S+R+A<30$ , then $S+R+A=10$ or $S+R+A=20$ .

If $\color{#3D99F6} S+R+A=10$ , then from tens place $1+R+M$ is divisible by $10$ . But $1+R+M<20$ , so $\color{#3D99F6} 1+R+M=10$ . Since $G>M$ there must be a carry over from hundreds place and since $1+A+M<20$ it must be $1$ . Thus $\color{#3D99F6} 1+A+M=10+R$ . Combine three blue relations to get $\color{#D61F06} A=2R$ and $\color{#D61F06} S=10-3R$ . Thus $R$ can be $1, 2, 3$ only and gives $S=7, 4, 1$ , $A=2, 4, 6$ , $M=8, 7, 6$ , and $G=9, 8, 7$ respectively. Notice that only $\color{#D61F06} (R, S, A, M, G)= (1, 7, 2, 8, 9)$ contains no repeated digit and it is a solution.

If $\color{#3D99F6} S+R+A=20$ , then from second column $2+R+M$ is divisible by $10$ . But $2+R+M<20$ , so $\color{#3D99F6} 2+R+M=10$ . Again, since $G>M$ one can conclude that $\color{#3D99F6} 1+A+M=10+R$ . Combine three blue relations to get $\color{#D61F06} A=2R+1$ and $\color{#D61F06} S=19-3R$ . Thus $R$ can only be $4$ and gives $S=7$ , $A=9$ , $M=4$ , and $G=5$ . There is a repeated digit here.

Therefore, $(R, S, {\color{#D61F06}A}, M, G)= (1, 7, {\color{#D61F06}2}, 8, 9)$ is the only solution.

For me it is absolutely beyond my intelligence limits to solve such digit puzzles honestly So I prefer to solve them writing nested loops with conditions using Python That time it took 20 lines of code I have split some lines for better readability

There definitely has to be a carryover in all the additions (except the one involving the thousands) since M <> G and R +M + 1/2/3 cannot be equal to A but would be of the form 10k + A. Now, it seems reasonable to assume the carry over in all the additions (except the additions involving the units digits : S, R A, and M) is 1.

Now S + R + M + A can be 30 at max if the digits are 9,8,7 and 6 in whatever order but this results in a digit other than the 4 and hence is not possible since the sum has to end in 1 of the 4 digits.

So,

S + R + A + M = 20 + M…..(!)

or

S + R + A + M = 10 + M…(2)

Let us examine (1)

Now M + 1 = G

A + M + 1 = 10 +R

A + M = 9 + R

R + A + M + 2 (carry over of 2 from the first addition) = 10 + A

2R + 11 = 10 + A

A – 2R = 1

A = 2R + 1

Giving the possibilities R=1,A=3 / R=2, A = 5/ R = 3, A = 7,/ R = 4, A = 9

Now S + R + A = 20

Which leaves only the last possibility of R = 4, A = 9,

This gives us S = 7.

To find the value of M, we use R + A + M + 2 = 10 + A

4 + 9 + 2 + M = 19

Giving a value of M = 4 but this equals R and hence is not possible

Now we go to (2)

A + M + 1 = 10 + R..

R + (A + M + 1) = 10 + A

R + 10 + R = 10 + A

10 + 2R = 10 + A

A = 2R

2R + M + 1 = 10 + R

R + M + 1 = 10

R + M = 9

Let us start with

R = 1, A = 2, M = 8

Which works !!

The remaining digits are S = 7 (S + R + A = 10) and G = M + 1 = 9

A + R + S = 10, ! + R + M = 10, 1 + A + M = R + 10. 1 + M = G,

R + G = 10. A + G = R + 10 = 2R + G, A = 2R, 3R + S = 10 = R + G

2R + S = G, A + S = G, R + G = 10, R = 1, G = 9. M= 8, S = 7, A = 2

                                 8   2   1   7

                                       8    2   1

                                              8    2

    +                                               8

                                     9     1    2     8

LaTex: $\Large \overline{MARS}+\overline{MAR}+\overline{MA}+\overline{M} = \overline{GRAM}$

LaTex: $\normalsize (1000M+100A+10R+S)+(100M+10A+R)+(10M+A)+(M) = 1000G+100R+10A+M$

LaTex: $\normalsize (0+0+0+S)+(0+0+R)+(0+A)+(M) \equiv 0+0+0+M \pmod {10}$

LaTex: $\Large R+S+A \equiv 0 \pmod {10} \iff A+S \equiv 9R \pmod {10}$

LaTex: $\large \text {We get too, adding in column :}$

LaTex: $\Large S+R+A+M=M \iff A+S \equiv 9R \pmod {10} \; (I)$

LaTex: $\Large 1+R+A+M=A \iff 9 \equiv M+R \pmod {10}\; (II)$

LaTex: $\Large 1+A+M=R \iff A+1 \equiv R-M \pmod {10} \; (III)$

LaTex: $\Large 1+M=G \iff \boxed{G-M=1} \iff G \equiv M+1 \pmod{10} \; (IV)$

LaTex: $\Large (II) \;\text {and}\; (III) \Rightarrow \color{#3D99F6}{ A \equiv 2R \pmod {10} }\; (V)$

LaTex: $\Large (I) \;\text {and}\; (V) \Rightarrow S \equiv 7R \pmod {10} \; (I)$

LaTex: $\Large \color{#3D99F6}{A \equiv 2R \pmod {10} \;\Rightarrow \text {(A is even)} \iff A \in \{2;4;6;8\} }$

Latex: $\Large if \;A=8 \Rightarrow \text{using} (I)(II)(III)(IV)(V) : A=S=8 \; \text {unacceptable}$

Latex: $\Large if \;A=6 \Rightarrow \text{using} (I)(II)(III)(IV)(V) : A=M=6 \; \text {unacceptable}$

Latex: $\Large if \;A=4 \Rightarrow \text{using} (I)(II)(III)(IV)(V) : A=S=4 \; \text {unacceptable}$

Latex: $\Large if \;A=2 \Rightarrow \text{using} (I)(II)(III)(IV)(V) : \; \text {we get : }$

Latex: $\Large \color{#624F41}{ \boxed{R=1}}$ Latex: $\Large \color{#624F41}{ \boxed{G=9}}$ Latex: $\Large \color{#624F41}{ \boxed{M=8}}$ Latex: $\Large \color{#624F41}{ \boxed{S=7}}$

Latex: $\Large \text {That solution meets the requirements of (I)(II)(III)(IV)(V), then : }$

Latex: $\Huge \color{#D61F06}{\boxed {A = 2}}$

at first, as each letter represents nonzero digit, so,we can say that all the letter represents between 1 to 9 (integers).

secondly,M,A,R,S,G-------so, there will be 5 integers among 1 to 9.

now,column 4 from left side------

S + R + A + M = 10 + M…(2)...........................[if they are 9,8 7,6 it would be more than 10, but it is not possible.]

now,

Now M + 1 = G

or, A + M + 1 = 10 +R

or, A + M = 9 + R

or, R + A + M + 2 ( 2 from the first addition) = 10 + A

or, 2R + 11 = 10 + A

or, A – 2R = 1

or, A = 2R + 1

now, assuming those possibilities,

we can say that,

A + M + 1 = 10 + R..

or, R + (A + M + 1) = 10 + A

or, R + 10 + R = 10 + A

or, 10 + 2R = 10 + A

or, A = 2R

or, 2R + M + 1 = 10 + R

or, R + M + 1 = 10

or, R + M = 9...................................[1]

so, at last counting all those possibilities ,we can say that,

A=2

S=7

G=9

now, from [1], we have got that R+M=9

so, only possible value for these two which fills the condition is, R= 8 or 1 , M=1 or 8. but R=8 and M=1 fills the requirement.

so ,we can say that,

M=8

A=2

R=1

S=7

G=9 ..............................this must be the solution.

Subtract 10A+M to simplify the problem to $\begin{matrix} M & A & R & S \\ & M & 0 & R \\ & & M & A \\ G & R & 0 & 0 \end{matrix}$ From the ones column we see $S+R+A=$ either $10$ or $20$ , so a 1 or 2 is carried to the tens column.

From the tens colums we then see $R+M =$ either $8$ or $9$ (Note that the sum of 2 different digits is 17 at maximum), so a 1 is carried to the hundreds column.

From the hundreds column we then have $1+A+M= R+10$ .

Substitute $M= \textrm{(8 or 9)} -R$ into this, we get $A= 2R+ \textrm{(0 or 1)} \Rightarrow R<5$

This leaves us with these possibilities $R \in\{1,2,3,4\}, M \in\{8-R, 9-R\}$ , where A,S and G are functions of R and M We check each possibility:

So the only correct solution is in the last line, and $A = \boxed{2}$

Since A plus M and a carry of 1 gets R, then going back to S+R+A+M, A+M must be greater than 10, and A is between 1 and 5, so whatever number it gives from 11 to 19 one must write the unit digits +1 as R, since M=R, and at the same time M must have a carry of 1. We know that M is not 9, then we can omit some cases, applying this cases we are left with 5 options to try, from which just the case of letting M=8, A=2, R=1+1, and S=7, is the only one that works, because S+R+A=10 in order to get S+R+A+M=M, then from that data we get G=9.... Sorry if I don't express my idea clearly, I swear I could explain it better face to face...

First column from left gives G = M+1,
Second column shows 1+A+M=10+R,
Or, A+M = 9+R, The sum results into following equation;
1000M+100A+10R+S+100M+10A+R+10M+A+M = 1000G+100R+10A+M,
Substituting G=M+1, and M=9+R-A, and simplifying gives,
9A = 21R +S - 10,
Or, A = 2R + (3R + S - 10)/9,
This equation shows that R can not exceed 4 as A is less than 9 (A=9 is also not possible as in A+M=R+9, M can not equal R).
For values of R less than or equal to 4,
(3R + S) will have to be equal to 10 so as to avoid values of A in fraction.
This gives A = 2R or R = A/2,
and 3R+S=10 or 3A+2S = 20,
Since A is not equal to S, A and S are not equal to 4.
Similarly, A+M=9+R gives,
A + 2M = 18, and since A and M are not equal, A and M can not be 6, so from 3A+2S=20, S can not be 1.
Now, from 3A + 2S = 20, we have,
A = 6 - (S - 1)/3, for A to be a natural number,
S should be 1, 4 or 7, but we have already seen above that S can not be 1 or 4,
so S=7 and consequently A = 2.

I used matrix logic by setting up a table with rows 1-9 and columns G R A M S.

I then used facts from the system of equations you can create to eliminate all possibilities. For instance, starting with the first possibility that the carry from ones column was 1, since the thousands place column has M as input in G is the output, and we know they are not allowed to be equal, then the hundreds place has at most a carry of one. This means G = M+1 and since M,G<>0 then G<>1 and M<>9 etc...

Other facts I found, A=2R so A is even S=3R R+M=9

I added other facts I found along the way eliminating down to only one poss I added other facts I found along the way. This illuminated down to only one possibility. A=2