For a positive integer $n$ and nonzero digits $a$, $b$, and $c$, let $A_n$ be the $n$-digit integer each of whose digits is equal to $a$; let $B_n$ be the $n$-digit integer each of whose digits is equal to $b$, and let $C_n$ be the $2n$-digit (not $n$-digit) integer each of whose digits is equal to $c$. What is the greatest possible value of $a + b + c$ for which there are at least two values of $n$ such that $C_n – B_n = A_n^2$?
What I have tried:
I start by trying $n = 1$ and $n = 2$. These give the system of equations $11c – b = a^2$ and $1111c – 11b = (11a)^2$. These imply that $a^2 = 9c$, so the possible $(a, c)$ pairs are $(9, 9)$, $(6, 4)$, and $(3, 1)$. The first puts $b$ out of range but the second makes $b = 8$. We now know the answer is at least $6 + 8 + 4 = 18$.
How should I continue?
Help is appreciated!
Also, if you are nice, can you PLEASE help me on this question($N$'s base-5 and base-6 representations, treated as base-10, yield sum $S$. For which $N$ are $S$'s rightmost two digits the same as $2N$'s?)?
Thanks!
Max0815
Best Answer
We note that $A_n = a\cdot\dfrac{10^n - 1}{9}$, $B_n = b\cdot\dfrac{10^n - 1}{9}$, and $C_n = c\cdot\dfrac{10^{2n} - 1}{9}$. Then, we have $c\cdot\dfrac{10^{2n} - 1}{9} - b\cdot\dfrac{10^n - 1}{9} = (a\cdot\dfrac{10^n - 1}{9})^2$, and when we multiply everything out and simplify, we get $$9c\cdot10^n + 9c - 9b = a^2 \cdot 10^n - a^2 \implies (9c-a^2) \cdot 10^n + 9c-9b + a^2 = 0$$ Now, note that since if two distinct values of $n$, say $x$ and $y$ work, then we get the equations$(9c-a^2) \cdot 10^x + 9c-9b + a^2 = 0$ and $(9c-a^2) \cdot 10^y + 9c-9b + a^2 = 0$, and subtracting gets $(9c-a^2) \cdot (10^y - 10^x) = 0$, and since $10^y - 10^x \neq 0$, we must have $9c = a^2$. Therefore, we get three solutions for $(a,c)$, which only $(6,4)$ gives an integer value of $b$, which means the solution $(a,b,c)$ is $(6,8,4) \implies 18$.
You were on the right track for finding $a^2 = 9c$, however, the problem does become easier after converting to $10^n$ instead of dealing with increasingly larger numbers; another way was to prove that $(a,c) = (3,1)$ is too small even if $b = 9$, and that $(a,c) = (9,9)$ has no solutions by noting that $C_n - (A_n)^2 = B_n$ and that $99 - 9^2 = 9(11-9) = 18$, $9999 - (99)^2 = 99(101-99) = 99 \cdot 2$, and that $9_n - (9_n)^2 = 2 \cdot 9_n$. (This can be proved by realizing that $C_n = A_n \cdot (A_n + 2)$.) Since $2 \cdot 9_n$ can never be equal to $B_n$, as $2 \cdot 9_n$ has a units digit of $8$ but it's first digit is $1$, we are done.