Greatest value of digits from adding numbers $\begin{array}
&&N&R\\
+&R&N\\\hline
A&B&C
\end{array}$
The addition problem above is correct. If N, R, A, B, and C are different digits, what is the greatest possible value of B+C?
is there an non-pluggining in way( maybe algebraic, geometric...) to prove that the greatest possible value of B+C=11?
I don't find plugging in numbers to solve this problem a good solution as I can't think of all the possible numbers. If possible, how do you plug in numbers and quickly determine that B+C is at most 11?
 A: I'm not convinced that there is a solution that doesn't involve any case checking, but here are some observations that will help:


*

*NR $= 10 N + R$ and RN $= 10R + N$ so $$\text{NR}+\text{RN}=11N +11R=11(N+R)$$

*ABC is divisible by $11$, and since $$\text{ABC}=100A + 10B + C \equiv A-B+C \pmod {11}$$we must have $A+C \equiv B \pmod{11}$, and hence since $0 \le A,C\le 9$, $$A+C=B$$

*To maximise $B+C$, we therefore wish to minimise $A$ and maximise $B$. We can check $A = 0$ is not possible, so $A \ge 1$. If $A = 1$, checking cases we see that $B=6$ and $C=5$.
A: No real way to avoid cases.
Since $B\neq C$ then $R+N=10+C$, and $B=1+C$, and $B+C=1+2C$.
If $C=6$ then the only unequal solution to $R+N=16$ is $R,N=7,9$ and then $B=7$, which means $7$ repeats. 
If $C=7$ then $R,N=8,9$ and then $B=8$, so $8$ repeats.
If $C>7$ then there is no pair of distinct $R,N$.
If $C=5$ then $R,N=78$ gives an example, and then $B=6$ and $B+C=11$.
A: "NR" in decimal notation is $10 N + R$, "RN" is $10 R + N$, so the sum is $11(R+N)$.  $R+N$ is the sum of two distinct nonzero digits, which  could be
anything from $3$ to $17$, but ABC must have three digits, so $R + N > 9$.
That leaves just 8 possibilities: 
$$\eqalign{11 \times 10 &= 110\cr
         11 \times 11 &= 121\cr
         11 \times 12 &= 132\cr
         11 \times 13 &= 143\cr
         11 \times 14 &= 154\cr
         11 \times 15 &= 165\cr
         11 \times 16 &= 176\cr
         11 \times 17 &= 187\cr}$$
But $N+R=17$ only happens with $8+9$ (in either order), and the $8$ conflicts with B.  Similarly $N+R=16$ only happens with $7+9$, and the $7$ conflicts.
$N+R=15$ happens with $6+9$ or $7+8$, and the latter provides a solution with $B+C=11$: $78 + 87 = 165$.
