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?

Is $ab$ a representation of the number? Like $ab = 23$ means $a = 2$ and $b = 3$? — Kaster, Aug 26 '15 at 23:26
It makes no sense to say $ab$ is a two-digit number and $abx$ is a single digit number ($2$) unless $x$ is a fraction. At a minimum you should clarify what kind of number $x$ is supposed to be (and resolve whether the claims $abx=2$ and $bc*x = 3$ are correctly stated). — hardmath, Aug 27 '15 at 00:02

score 2 · Answer 1 · answered Aug 26 '15 at 23:41

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.

Brian M. Scott · Accepted Answer · 2015-08-27T00:19:02.963

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 $20\cdot 29=580$.

edited Aug 27 '15 at 00:19

answered Aug 26 '15 at 23:47

Brian M. Scott

616,228

$x = 1/23$ and $1/29$, not $23$ and $29$ (at least according to the terms of the problem). – Brian Tung Aug 27 '15 at 00:07
@Brian: Oops; misremembered what I was doing. Thanks! – Brian M. Scott Aug 27 '15 at 00:20
It was rather confusing. I would rather have seen $ab/x = 2, bc/x = 3, (a+b+c)x =$ ? – Brian Tung Aug 27 '15 at 00:27

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

2 Answers2