Determine all three-digit positive integers abc (a is hundreds b tens and c ones) such that 8abc = 3cba.
Here is my how I have started:
8 (100a + 10b + c) = 3(100c + 10b + a)
800a + 80b + 8c = 300c + 30b + 2a
797a + 50b – 292c = 0
50b = 292c – 797a
b = (292c – 797a)/50
Since abc and cba both need to be even we know that a must be even but not c because factor 292 makes the product even.
Then I started to test different values for a and c, it didn't take so long because a can be larger than 2, so the only values I tested were 1 and 2 for a. For c I tested all values up to 7 (then I found the answer) and it was kind of easy because you don't need to do the whole calculation since if the unit digit of the product of 292 multiplied by c minus 4 and 7 is not 0 we know that the difference is not divisible with 50.
So the only answer I found is c = 7 and a = 2 so b = 9
But I can't prove it algebraically or write a better solution for this problem, I don't think testing is the best method for solving it…
Best Answer
$$797a+50b=292c$$ Here we see that $a$ must be even, and if $a\ge4$ then the LHS is too big for the RHS to match in magnitude (remember that $0\le c\le 9$). Hence $a=2$: $$1594+50b=292c$$ Now we see that $c$ must be at least $6$ so that the RHS can match the LHS in magnitude, and must be $2$ or $7$ to match in the last digit (the LHS's last digit must be $4$). Hence $c=7$, from which we get $b=9$ and the unique solution as $\overline{abc}=297$.