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Difference between revisions of "PaperMath’s sum"
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== PaperMath’s sum==
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== 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>\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}=(\sum_{j=0}^{n -1}{(3 \times 10^j)})^2 + \sum_{k=0}^{n-1} {(2 \times 10^k)}</math>
  
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==Proof==
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== Proof ==
  
 
First, note that the <math>x^2</math> part is trivial multiplication, associativity, commutativity, and distributivity over addition,
 
First, note that the <math>x^2</math> part is trivial multiplication, associativity, commutativity, and distributivity over addition,
  
 
Observing that  
 
Observing that  
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<math>\sum_{i=0}^{n-1} {10^i} =  
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<math>\sum_{i=0}^{n-1} {10^i} = \frac{10^{n}-1}{9}</math> and <math>(10^{2n}-1)/9 = 9((10^{n}-1)/9)^2 + 2(10^n -1)/9</math> concludes the proof.
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(10^{n}-1)/9</math>
 
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and  
 
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<math>(10^{2n}-1)/9 = 9((10^{n}-1)/9)^2 + 2(10^n -1)/9</math>
 
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concludes the proof.
 
  
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==Problems==
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== Problems ==
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AMC 12A Problem 25
 
  
 
For a positive integer <math>n</math> and nonzero digits <math>a</math>, <math>b</math>, and <math>c</math>, let <math>A_n</math> be the <math>n</math>-digit integer each of whose digits is equal to <math>a</math>; let <math>B_n</math> be the <math>n</math>-digit integer each of whose digits is equal to <math>b</math>, and let <math>C_n</math> be the <math>2n</math>-digit (not <math>n</math>-digit) integer each of whose digits is equal to <math>c</math>. What is the greatest possible value of <math>a + b + c</math> for which there are at least two values of <math>n</math> such that <math>C_n - B_n = A_n^2</math>?
 
For a positive integer <math>n</math> and nonzero digits <math>a</math>, <math>b</math>, and <math>c</math>, let <math>A_n</math> be the <math>n</math>-digit integer each of whose digits is equal to <math>a</math>; let <math>B_n</math> be the <math>n</math>-digit integer each of whose digits is equal to <math>b</math>, and let <math>C_n</math> be the <math>2n</math>-digit (not <math>n</math>-digit) integer each of whose digits is equal to <math>c</math>. What is the greatest possible value of <math>a + b + c</math> for which there are at least two values of <math>n</math> such that <math>C_n - B_n = A_n^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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==Notes==
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([[AMC 12A Problem 25|Source]])
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Papermath’s sum was named by the aops user Papermath, and is a very fantastic formula
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== Notes ==
 +
 
 +
Papermath’s sum was named by the aops user Papermath.
 +
 
 
==See also==
 
==See also==
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*[[Cyclic sum]]
 
*[[Cyclic sum]]
 
*[[Summation]]
 
*[[Summation]]
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[[Category:Algebra]]
 
[[Category:Algebra]]
 
[[Category:Theorems]]
 
[[Category:Theorems]]
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{{Stub}}

Latest revision as of 14:15, 2 February 2025

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.

See also

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