How many 3-digit numbers can be multiplied by $\frac{4}{3}$ and end up as the same 3-digit number with the last digit moving to the front?

$\overline{ABC}\cdot \frac { 4 }{ 3 } = \overline{CAB}$

**
Note:
**
In this problem,
$\overline { ABC }$
is defined as
$A$
being the first digit,
$B$
the second, and
$C$
the third, not
$A\cdot B\cdot C$
.

8
1
10
4
0
2
6
9

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Rather than working with the three single-digit numbers $A$ , $B$ and $C$ , I'm going to only consider two numbers: $a$ , which equals the first two digits, and $b$ , which equals the last digit. This simplifies the equation to: $\frac{4}{3}(10a+b)=100b+a\\ 40a+4b=300b+3a\\ 37a=296b\\ a=8b$ We know that $b$ has to be only a single digit, so we just need to consider all the values of $b$ such that $8b$ has 2 digits. All of these must be less than 3 digits, and $b=0$ and $b=1$ give single-digit values of $a$ and don't count, so we are left with $\boxed8$ available options for $b$ .