67 of 100: Tram O'Rama

A+R = A must be true when R=0 or there is a 1 at hand. R is non-zero.
So, A+R+1= A+10. or, R=9.

Now, T+M+1 = 9. Or, T+M= 8. So, A=8. Now it shapes as:

At tens digit, 9+8= 17, and we have noting at hand. So, M=7. Putting M=7 and T=1, we get a correct answer:

So, M=7

We note that the thousands column is $M+T+{\color{#D61F06}\{_0^1}=R$ , where $\color{#D61F06}\{_0^1$ denotes the carry-forward of 1 or 0. This means that $M+T + \{_0^1 \le 9$ . We also note that the units column is also $T+M=A$ . This means that $T+M \le 8$ and $M+T+{\color{#D61F06}1} = R$ , implying that $R=A+1$ . Also implying that in the hundreds and tens columns $A+R=R+A>10$ . $\implies A+R+1=10+A$ $\implies R=9$ , $\implies A=R-1=8$ . The tens column $R+A=10+M$ $\implies 9+8 = 10+M$ $\implies M = \boxed{7}$ . And the units column $T+M=A$ $\implies T+7=8$ $\implies T=1$ .

$\begin{array} {ccccc} & \color{#D61F06}\boxed{7} & \color{#20A900}8 & \color{#3D99F6}9 & \color{#69047E}1 \\ + & \color{#69047E}1 & \color{#3D99F6}9 & \color{#20A900}8 & \color{#D61F06}\boxed{7} \\ \hline & \color{#3D99F6}9 & \color{#20A900}8 & \color{#D61F06}\boxed{7} & \color{#20A900}8 \end{array}$

notice that a maximum of 1 can carry over from the third digit to the second digit. so either $A+R\equiv A \mod{10}\to R\equiv 0 \mod{10}\to R=0 \\ \text{or}\\A+R+1\equiv A\mod{10}\to R\equiv -1\equiv 9 \mod{10} \to R=9$ since R is nonzero, the first option is impossible meaning $R=9$ . next we see a one carries over the first digit, meaining $M+T+1=R\to M+T+1= 9\to M+T=8$ from the fourth digit $T+M\equiv A\mod{10}\to 8=A$ . now the third digit says $R+A\equiv M \mod{10} \to M\equiv 17 \mod{10}\to M=\boxed{7}$ and ofcourse $T=8-M=1$ . the crypto gram is $\begin{array} {ccccc} \large & & \color{#D61F06}7 & \color{#20A900}8 & \color{#3D99F6}9 & \color{#69047E}1\\ \large + & &\color{#69047E}1 & \color{#3D99F6}9& \color{#20A900}8 & \color{#D61F06}7 \\ \hline \large & & \color{#3D99F6}9 & \color{#20A900}8& \color{#D61F06}7 & \color{#20A900}8 \end{array}$

A straightforward solution in Javascript Outputs: 0 and 7

From the hundreds column with the digit A in the sum and also being the answer means that there must be a 'carry over' from the previous column so that 1 + R = 10. So R = 9. This 'carry over' + R to 'carry over' 1 on the thousands column gives that 1 + ( M + T) = R (=9). So M + T = 8 (= A from the 1sr column). And the first column did not have a carry over of 1. So from the 2nd column R + A is 9 + 8 = 17. So the M = 7 and there's 1 to carry over from column 2, which we knew already. And to get the last digit just for completeness from the 1st column again M + T = A, and M = 7, A = 8 gives T = 1. So the sum is 7,891 + 1,987 = 9,878 which is correct. And M has been seen to be 7. Regards, David

$T + M = A \\ T = M - A \\ \\ R + A = 10 + M \\ R + T + M = 10 + M \\ R + T = 10 \\ \\ A + R + 1 = 10 + A \\ R + 1 = 10 \\ R = 9 \\ \\ T = 1 \\ \\ M + T + 1 = R \\ M + T + 1 = 9 \\ M + T = 8 \\ \\ M + T = A \\ A = 8 \\ \\ M + 1 = 8 \\ \large M = \boxed{\color{#D61F06}7}$

from, the left most column, we get= $T+M=A$ ....................[suppose]

again, from the right most column,we get= $M+T=R$ ............[suppose]

we know that, $A+B=B+A$

so, we can say that $M+T≠R$ as after doing some calculation ,it comes at the last.and, we can assume that, $M+T=A$

and now,, $A+R+1=10+A$

or, $R=9$

again,M+T+1=9

or, $M+T=A=8$

so, just doing the sum, $R+A=9+8=17$

we have to take 7 for M and 10 for the next calculation.so, its clear that, $M=7$

I used arithmetic modulo 11 to help me. Here is how. From the leftmost and rightmost column we infer $R = A + 1$ (also see Arjen's solution). From the second column and $R \ne 0$ , we infer $R = 9$ and there was a carry from the third column. Hence, $A = 8$ . Taking everything ${}\bmod 11$ , we find $M - A + R - T + T - R + A - M = R - A + M - A \pmod{11}$ . This simplifies to $0 = R -2A + M \pmod{11}$ . Using $R, A = 9, 8$ , we find $M = 7 \pmod{11}$ , which implies $\fbox{M = 7}$ .

Compare the first and fourth columns. In both, $T$ and $M$ are added together, but they result in a different digit. This difference must be due to the fact that there is a carry-over from the second column to the first. Thus $T + M = A,\ \ T + M + 1 = R,\ \ R = A + 1.$ In the same way, the second and third column show that $R + A = 10 + M,\ \ R + A + 1 = 10 + A,\ \ A = M + 1.$ Since $T + M = A$ and $1 + M = A$ , we know that $T = 1$ .

From the second column we also see that $A + R + 1 = A + 10\ \ \ \therefore\ \ \ \ R = 9.$ We immediately conclude that $A = 8$ and $\boxed{M = 7}$ .

A little investigation in the addition reveals that:

T + M = A, R + A = 10 + M 10 – T = R, 1 + A + R = 10 + A, R = 9, T = 1

1 + M + 1 = 9, M = 7, A = 8.

Therefore, M = 7.

Now, is that really so "spiky"?

To get past two cacti shouldn't the solution require something that is more than one level deep?

Since addition of two four-digit numbers here yields also four-digit number, we can be sure that $M+T+1<10$ , and hence $M+T<10$ also. Then, we can with certainty write $M+T=A$ . From that we can see that $M<A$ and thus $R+A=10+M$ . Now, $\begin{aligned} & A+R+1=10+A \\ & \rightarrow R=9\end{aligned}$ .

Next, $\begin{aligned} & M+T+1=9 \\ & \rightarrow M+T=A=8 \end{aligned}$ .

From $R+A=9+8=10+R$ , we derive $M=7$ .

1) Since A +R = M (hundreds) and R + A = A (tens), only one of these columns has a carry-over one from the prior column (tens or ones).

2) Since R is a digit (def: less than 10) and R+A=A there must be a carry over from the tens, such that R+1=10. Now we know R is 9.

3) Since R+A=M and we k ow there is no carry-over from the ones, M=A-1.

4) From the hundreths see T + M = R and from the ones : T+M=A. We know the ones do not have a carry over, so: T+M+1=A+1=R=9. That makes A=8.

Result: M=A-1=7

In the second column, A+R=A ,.
So, it is sure that, A+1=R.
Now, maximum value of (M+T)=8,as then A=8 and R=9.
In the third column,R+A=M,.
So, it is sure that, M+1=A.
So, T=1.
Now, from third column, (M+2)+(M+1)=M;i.e.(M+2)+(M+1)=M+10
or,M=7
So, the answer is 7

M + T= A

M + T + 1 = R

R = A+ 1

Looking at the third addition from the right, 2A + 2 = 10 + A (There is definitely a carryover in the R + A addition since M is smaller than A & R. So, A + R + 1 = 10 + A and R = A + 1)

A = 8

R= 9

M = 7

T= 1

Choose left = min {all answers}

Choose right = max {all answers}

Choose mid = (left+right)/2

So there mid=7

Answer mid