How to find $\frac{a+b+c}x$?

Question

$ab$ and $bc$ are two digit numbers. if $ab*x=2 $ and $bc*x=3$ then find $\frac{a+b+c}x$. (* is multiplication)

It looks simple but I couldnt go further. $$17b=2(15a-c)\iff b\mid2 \quad and\quad 17\mid (15a-c)$$

a=4 b=6 and c=9 suffice this. Is there a way without trying numbers?

Answer 1

You know that $30a=17b+2c$ and hence that $b$ is even. Setting $b=2d$, we have $15a=17d+c$, where $d\in\{1,\ldots,4\}$. Since there are only four possibilities for $d$, trial and error is the way to go.

• If $d=1$, we have $15a=17+c$, which has no solution with single digit $c$.
• If $d=2$, we have $15a=34+c$, with the same problem.
• If $d=3$, we have $15a=51+c$, which has the solution $a=4,c=9$.
• If $d=4$, we have $15a=68+c$, which has the solution $a=5,c=7$.

Thus, either $a=4,b=6$, and $c=9$, or $a=5,b=8$, and $c=7$. In the first case $x=\frac1{23}$, and in the second $x=\frac 1{29}$, so $\frac{a+b+c}x$ is either $19\cdot 23=437$ or $20\cdot 29=580$.

Answer 2

You're most of the way there. You've observed that $b$ is even. Now consider that $0 \leq c \leq 9$, so $17b$ must be no more than $18$ less than a multiple of $30$.

Both $17 \times 2 = 34$ and $17 \times 4 = 68$ fail to qualify, so that leaves $17 \times 6 = 102$ and $17 \times 8 = 136$. These two yield $a = 4, b = 6, c = 9$ and $a = 5, b = 8, c = 7$, respectively. Other than the degenerate solution $a = b = c = 0$, these are the only two solutions.