3-digit cyclics

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 $b$ .

We can rewrite the problem as $\frac { 4 }{ 3 } (100A+10B+C)=100C+10A+B$ , which can be simplified to $370A+37B=296C$ .

Because $\frac { 296 }{ 37 } =8$ , we can divide both sides by 37 to get $10A+B=8C$ .

Keep in mind that A, B, and C have to be positive integers less than 10. That's important because it means that in the left side of the previous equation, the tens place is controlled only by the value of A, and the ones place is controlled only by the value of B. So, to find solutions to the problem, we can input different values of C in hopes of getting values of A and B such that $\frac { 4 }{ 3 } 100A+10B+C$ is a 3-digit number (remember that $10A+B=8C$ ).

It turns out that there's 8 different solutions: 162, 243, 324, 405, 486, 567, 648, and 729.

Interestingly, when the solution 243 is multiplied by $\frac{4}{3}$ , you end up with 324, which is another solution to the problem!

Also interestingly, the largest solution 729 is actually ${ 9 }^{ 3 }$ . I would recommend that you spend some time trying to figure out why this is true - it will give you a deeper understanding of why the above process works. I also want you to think about why the fraction is $\frac{4}{3}$ . What do you think simplifies to $\frac{4}{3}$ ?

Mathematica

Select[Range[100,999],IntegerDigits[#*4/3]==RotateRight@IntegerDigits@#&]

{162, 243, 324, 405, 486, 567, 648, 729}