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PaperMath’s sum: Difference between revisions

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==AntandMonkeyDoctor’s sum==
PaperMath's sum is a thereom discovered by the AoPS user Papermath on October 8th, 2023.
1+1=2


==Proof==
== Statement ==
We will first prove a easier variant of PaperMath’s sum,
'''PaperMath’s sum''' states,


<math>\sum_{i=0}^{2n} {(9 \times 10^i)}=(\sum_{j=0}^n {(9 \times 10^j)})^2 + 9\sum_{k=0}^n {(2 \times 10^k)}</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>


This is the exact same as
Or
 
<math>\sum_{i=0}^{2n} {10^i}=(\sum_{j=0}^n {(3 \times 10^j)})^2 + \sum_{k=0}^n {(2 \times 10^k)}</math>
 
But everything is multiplied by <math>9</math>.
 
Notice that this is the exact same as saying
 
<math>\underbrace {9999\dots}_{2n}=(\underbrace {99\dots}_{n})^2+9(\underbrace {22\dots}_{n})</math>
 
Notice that <math>9(\underbrace {22\dots}_{n})=2(\underbrace {99\dots}_{n})</math>
 
Substituting this into <math>\underbrace {9999\dots}_{2n}=(\underbrace {99\dots}_{n})^2+9(\underbrace {22\dots}_{n})</math> yields
<math>\underbrace {9999\dots}_{2n}=(\underbrace {99\dots}_{n})^2+2(\underbrace {99\dots}_{n})</math>
 
Adding <math>1</math> on both sides yields
 
<math>10^{2n}= (\underbrace {99\dots}_{n})^2+2(\underbrace {99\dots}_{n})+1</math>
 
Notice that <math>(\underbrace {99\dots}_{n})^2+2(\underbrace {99\dots}_{n})+1=(\underbrace {99\dots}_{n}+1)^2=(10^n)^2=10^{2n}</math>
 
As you can see,
<math>\sum_{i=0}^{2n} {(9 \times 10^i)}=(\sum_{j=0}^n {(9 \times 10^j)})^2 + 9\sum_{k=0}^n {(2 \times 10^k)}</math>
 
Is true since the RHS and LHS are equal
 
This equation holds true for any values of <math>n</math>. Since this is true, we can divide by <math>9</math> on both sides to get


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


And then multiply both sides <math>x^2</math> to get
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} {(x^2 \times 10^i)}=(\sum_{j=0}^n {(3x \times 10^j)})^2 + \sum_{k=0}^n {(2x^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>


Or
== Proof ==


<math>x^2\sum_{i=0}^{2n} {10^i}=(x \sum_{j=0}^n {(3 \times 10^j)})^2 + x^2\sum_{k=0}^n {(2 \times 10^k)}</math>
First, note that the <math>x^2</math> part is trivial multiplication, associativity, commutativity, and distributivity over addition,


Which proves PaperMath’s sum
Observing that
<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.


==Problems==
== Problems ==
2018 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>


==Notes==
([[2018 AMC 10A Problems/Problem 25|Source]])


PaperMath’s sum was discovered by the aops user PaperMath, as the name implies. It was stolen from Sagaman.
==See also==


==See also==
*[[Cyclic sum]]
*[[Cyclic sum]]
*[[Summation]]
*[[Summation]]
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[[Category:Algebra]]
[[Category:Algebra]]
[[Category:Definition]]
[[Category:Theorems]]
{{Stub}}

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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