# What is the greatest possible value of $a + b + c$ for which there are at least two values of $n$ such that $C_n - B_n = A_n^2$?

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

• How do you get from your first two equations to $a^2=9c?$ You can divide the second by $11$ to get $101c-b=11a^2$ Commented Feb 5, 2019 at 23:53
• I'm not sure what you mean by your first equation. And how does dividing by 11 help though... :( I'm confused Commented Feb 5, 2019 at 23:55
• You said correctly that for $n=1$ you have $11c-b=a^2$ and for $n=2$ you have $1111c-11b=(11a)^2$. I don't see how either of those leads to $a^2=9c$. I divided the $n=2$ equation by $11$ to get my result, which I think is a simpler equation to work with. Commented Feb 5, 2019 at 23:59
• Right... ahh I think I had an algebra error. Let me try again with ur equation... Commented Feb 6, 2019 at 0:02
• @RossMillikan Now subtract $11c - b = a^2$ from $101c - b = 11a^2$. Doesn't it give $90c = 10a^2$? Commented Feb 6, 2019 at 0:03

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.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 62, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://math.stackexchange.com/questions/3101848/what-is-the-greatest-possible-value-of-a-b-c-for-which-there-are-at-least/3101908", "fetch_time": 1726614411000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-38/segments/1725700651835.53/warc/CC-MAIN-20240917204739-20240917234739-00414.warc.gz", "warc_record_offset": 349776878, "warc_record_length": 38561, "token_count": 1362, "char_count": 3622, "score": 4.15625, "int_score": 4, "crawl": "CC-MAIN-2024-38", "snapshot_type": "latest", "language": "en", "language_score": 0.9230234622955322, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00000-of-00064.parquet"}