Mirrored Cryptogram

Logic Level 3

$\large \begin{matrix} & A & B & C \\ & B & B & C \\ + & C & C & C \\ \hline \\ \ \\ \ \end{matrix} \qquad \quad \begin{matrix} & C & B & A \\ & C & B & B \\ + & C & C & C \\ \hline \\ \ \\ \ \end{matrix}$

$A$ , $B$ , and $C$ in the two cryptograms above are distinct digits from 1 to 9. If the two cryptograms sum up to the same number, what is the smallest possible sum of each?

The answer is 646.

1 solution

Chew-Seong Cheong
Jul 28, 2017

Equating the sums of the two cryptograms, we have:

$\begin{aligned} 100A + 120B + 113C & = 311C + 21B + A \\ 99A + 99B & = 198C & \small \color{#3D99F6} \text{Dividing both sides by }99 \\ A+B & = 2C \end{aligned}$

Since we want to get the smallest sum, we work with the smallest $A$ , $B$ and $C$ possible. And from $A+B=2C$ , we have $A$ or $B$ equal to 1 or 3 and $C=2$ , so that $1+3=2(2)$ . Since the sum $S = 100A+120B+113C$ , $A$ is multiplied by 100 while $B$ by 120, $S$ is smallest when $A=3$ , $B=1$ and $C=2$ . Then $S=300+120+226=\boxed{646}$ .

How do you know that $S$ is minimized when $(A,B,C) = (3,1,2)$ ? Did you try all $9\times9\times9=729$ cases?

Pi Han Goh - 3 years, 10 months ago

Mirrored Cryptogram

The answer is 646.

1 solution

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