Anomalous Cancellation

Using Garrett's notation $\overline{AB}=10a+b$ , $\overline{BC}=10b+c$ we get $\frac{10a+b}{10b+c}=\frac{a}{c} \\ b(10a-c)=9ac$ Note that $a$ , $b$ and $c$ are nonzero digits. $c=a$ is our trivial case, for $a \neq c$ let $d=c-a$ . Its clear that $d$ cannot take $\pm9$ . Equation becomes $b(9a-d)=9a(a+d) \\ 9a^2+9a(d-b)+bd=0$ It's immediate that $9|bd$ and since $d\neq \pm9$ only possible values for $b$ are $3, 6, 9$ .

For $b=3$ $c=\frac{10a}{3a+1}$ and we must have $3a+1|10(3a+1)-3(10a)=10$ , which gives $a=3$ only and this is one of trivial solutions.

For $b=6$ $c=\frac{20a}{3a+2}$ and we must have $3a+2|20(3a+2)-3(20a)=40$ , which gives $a=1, 2, 6$ and leads to $(a, b, c)=(1, 6, 4), (2, 6, 5)$ and trivial solution $(6, 6, 6)$ .

For $b=9$ $c=\frac{10a}{a+1}$ and we must have $a+1|10(a+1)-10a=10$ , which gives $a=1, 4, 9$ and leads to $(a, b, c)=(1, 9, 5), (4, 9, 8)$ and trivial solution $(9, 9, 9)$ .

We are done.

Clearly we have the 9 trivial solutions where $A=B=C$ , for example $\frac{33}{33}=\frac{3}{3}=1$ .

Now to find the non-trivial solutions. We start by replacing $\overline{AB}=10a+b$ and $\overline{BC}=10b+c$ :

$\frac{10a+b}{10b+c}=\frac{a}{c}$

$c(10a+b)=a(10b+c)$

$10ac+bc=10ab+ac$

$b(10a-c)=9ac$

$b=\frac{9ac}{10a-c}$

Since $b$ is an integer, we know that $(10a-c)\mid 9ac$ . For $1\leq a,c\leq9$ , we get the following triples $(a,b,c)$ :

$(1,6,4)\Longrightarrow\frac{16}{64}=\frac{1}{4}$

$(1,9,5)\Longrightarrow\frac{19}{95}=\frac{1}{5}$

$(2,6,5)\Longrightarrow\frac{26}{65}=\frac{2}{5}$

$(4,9,8)\Longrightarrow\frac{49}{98}=\frac{4}{8}=\frac{1}{2}$

There are four more triples, $(1,21,7)$ , $(1,36,8)$ , $(1,81,9)$ & $(2,12,8)$ , but since $1\leq b\leq9$ , we must disregard these solutions.

Therefore we have 4 non-trivial solutions and 9 trivial solutions, giving us our final answer of $\boxed{13}$ .

It is not clear from the wording of the question that we need to ignore the cases where A = B = 0, in which case the answer would increase from 13 to 22. . Could you please clarify this in the problem statement?

Discard the degenerate solutions where A = B = 0, because we want two-digit numbers without leading zeros. And discard C=0 because it causes a division by 0. So below we will assume A, B and C to be nonzero digits.

$\frac{\overline{AB}}{\overline{BC}} = \frac{A}{C} \Leftrightarrow \frac{10A+B}{10B+C}=\frac{A}{C} \Leftrightarrow 10AC+BC=10AB+AC$ $\Leftrightarrow 9AC=B(10A-C)$

Observe that $B(A-C) \equiv 0 \pmod 9$ . So $A=C \vee B=9 \vee (3|B \wedge 3|(A-C))$

If $A=C \neq 0$ , then the fraction $A/C$ evaluates to $1$ implying $\overline{AB}=\overline{BC}$ so that $A=B=C$

If $B=9$ then $AC=10A-C \Rightarrow C=10A/(A+1)$ leading to $(A,C)\in \{(1,5)(4,8)(9,9)\}$

If $B=3$ then $3AC=10A-C \Rightarrow C=10A/(3A+1)$ leading to $(A,C) \in \{(3,3)\}$

If $B=6$ then $3AC=20A-2C \Rightarrow C=20A/(3A+2)$ leading to $(A,C) \in \{(1,4)(2,5)(6,6)\}$

Summarized, we found $\boxed{13}$ solutions: $\frac{11}{11}=\frac11, \frac{22}{22}=\frac22, ... , \frac{99}{99}=\frac99, \frac{19}{95}=\frac15, \frac{49}{98}=\frac48, \frac{16}{64}=\frac14, \frac{26}{65}=\frac25$

I know none of you look for this but I am going to post it anyway.

for i in range(1, 10):
    for j in range(1, 10):
        for k in range(1, 10):
            if (i * 10 + j) / (j * 10 + k) == i / k:
                print(i, j, k)

Simple. Concise. Adaptive.

for(int a=1;a<10;a++)

{for(int b=1;b<10;b++)

{ for(int c=0;c<10;c++)

{if(9 a c+b c-10 a*b==0)

System.out.println("("+a+","+b+")  ("+b+","+c+")");
}
}
}

WRITE THIS CODING ON THE ACTION PERFORMED EVENT OF A BUTTON IN NETBEANS. U GET THE RESULT.

(1,1) (1,1) (1,6) (6,4) (1,9) (9,5) (2,2) (2,2) (2,6) (6,5) (3,3) (3,3) (4,4) (4,4) (4,9) (9,8) (5,5) (5,5) (6,6) (6,6) (7,7) (7,7) (8,8) (8,8) (9,9) (9,9)