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PaperMath’s sum: Difference between revisions

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PaperMath's sum is a thereom discovered by the AoPS user Papermath on October 8th, 2023.
== Statement ==
== Statement ==
'''Papermath’s sum''' states,
'''PaperMath’s sum''' states,


<math>\sum_{i=0}^{2n-1} {(10^ix^2)}=(\sum_{j=0}^{n-1}{(10^j3x)})^2 + \sum_{k=0}^{n-1} {(10^k2x^2)}</math>
<math>\sum_{i=0}^{2n-1} {\left(10^ix^2\right)}=\left(\sum_{j=0}^{n-1}{\left(10^j3x\right)}\right)^2 + \sum_{k=0}^{n-1} {\left(10^k2x^2\right)}</math>


Or
Or


<math>x^2\sum_{i=0}^{2n-1} {10^i}=(3x \sum_{j=0}^{n-1} {(10^j)})^2 + 2x^2\sum_{k=0}^{n-1} {(10^k)}</math>
<math>x^2\sum_{i=0}^{2n-1} {10^i}=\left(3x \sum_{j=0}^{n-1} {\left(10^j\right)}\right)^2 + 2x^2\sum_{k=0}^{n-1} {\left(10^k\right)}</math>


For all real values of <math>x</math>, this equation holds true for all nonnegative values of <math>n</math>. When <math>x=1</math>, this reduces to
For all real values of <math>x</math>, this equation holds true for all nonnegative values of <math>n</math>. When <math>x=1</math>, this reduces to


<math>\sum_{i=0}^{2n-1} {10^i}=(\sum_{j=0}^{n -1}{(3 \times 10^j)})^2 + \sum_{k=0}^{n-1} {(2 \times 10^k)}</math>
<math>\sum_{i=0}^{2n-1} {10^i}=\left(\sum_{j=0}^{n -1}{\left(3 \times 10^j\right)}\right)^2 + \sum_{k=0}^{n-1} {\left(2 \times 10^k\right)}</math>


== Proof ==
== Proof ==
Line 25: Line 27:
<math>\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20</math>
<math>\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20</math>


([[AMC 12A Problem 25|Source]])
([[2018 AMC 10A Problems/Problem 25|Source]])
 
== Notes ==
 
Papermath’s sum was named by the aops user Papermath.


==See also==
==See also==

Latest revision as of 15:51, 24 October 2025

PaperMath's sum is a thereom discovered by the AoPS user Papermath on October 8th, 2023.

Statement

PaperMath’s sum states,

$\sum_{i=0}^{2n-1} {\left(10^ix^2\right)}=\left(\sum_{j=0}^{n-1}{\left(10^j3x\right)}\right)^2 + \sum_{k=0}^{n-1} {\left(10^k2x^2\right)}$

Or

$x^2\sum_{i=0}^{2n-1} {10^i}=\left(3x \sum_{j=0}^{n-1} {\left(10^j\right)}\right)^2 + 2x^2\sum_{k=0}^{n-1} {\left(10^k\right)}$

For all real values of $x$, this equation holds true for all nonnegative values of $n$. When $x=1$, this reduces to

$\sum_{i=0}^{2n-1} {10^i}=\left(\sum_{j=0}^{n -1}{\left(3 \times 10^j\right)}\right)^2 + \sum_{k=0}^{n-1} {\left(2 \times 10^k\right)}$

Proof

First, note that the $x^2$ part is trivial multiplication, associativity, commutativity, and distributivity over addition,

Observing that $\sum_{i=0}^{n-1} {10^i} = \frac{10^{n}-1}{9}$ and $(10^{2n}-1)/9 = 9((10^{n}-1)/9)^2 + 2(10^n -1)/9$ concludes the proof.

Problems

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 $C_n - B_n = A_n^2$?

$\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20$

(Source)

See also

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