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Difference between revisions of "PaperMath’s sum"

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PapreMath's sum is a thereom discovered by the APS user Papermath on October 8th, 2023.
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== Statement ==
 
== Statement ==
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'''Papermath’s sum''' states,
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'''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} {(10^ix^2)}=(\sum_{j=0}^{n-1}{(10^j3x)})^2 + \sum_{k=0}^{n-1} {(10^k2x^2)}</math>
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<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>
  
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([[AMC 12A Problem 25|Source]])
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([[2018 AMC 10A Problems/Problem 25|Source]])
  
 
== Notes ==
 
== Notes ==
  
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Papermath’s sum was named by the aops user Papermath.
+
PaperMath’s sum was named by the aops user PaperMath.
 +
PaperMath is very orz.
  
 
==See also==
 
==See also==

Latest revision as of 22:28, 22 June 2025

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

Statement

PaperMath’s sum states,

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

Or

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

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

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

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)

Notes

PaperMath’s sum was named by the aops user PaperMath. PaperMath is very orz.

See also

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