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Difference between revisions of "PaperMath’s sum"

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==PaperMath’s sum==
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== PaperMath’s sum==
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This is a summation identities for decomposition or reconstruction of summations. Papermath’s sum states,
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This is a summation identities for decomposition or reconstruction of summations. PaperMath’s sum states,
 
  
 
<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} {(x^2 \times 10^i)}=(\sum_{j=0}^n {(3x \times 10^j)})^2 + \sum_{k=0}^n {(2x^2 \times 10^k)}</math>
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==Proof==
 
==Proof==
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We will first prove a easier variant of PaperMath’s sum,
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We will first prove a easier variant of Papermath’s sum,
  
 
<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} {(9 \times 10^i)}=(\sum_{j=0}^n {(9 \times 10^j)})^2 + 9\sum_{k=0}^n {(2 \times 10^k)}</math>
  
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This is the exact same as  
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This is the exact same as
  
 
<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>\sum_{i=0}^{2n} {10^i}=(\sum_{j=0}^n {(3 \times 10^j)})^2 + \sum_{k=0}^n {(2 \times 10^k)}</math>
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Notice that <math>9(\underbrace {22\dots}_{n})=2(\underbrace {99\dots}_{n})</math>
 
Notice that <math>9(\underbrace {22\dots}_{n})=2(\underbrace {99\dots}_{n})</math>
  
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Substituting this into <math>\underbrace {9999\dots}_{2n}=(\underbrace {99\dots}_{n})^2+9(\underbrace {22\dots}_{n})</math> yields  
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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>
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<math>\underbrace {9999\dots}_{2n}=(\underbrace {99\dots}_{n})^2+2(\underbrace {99\dots}_{n})</math>
 
  
 
Adding <math>1</math> on both sides yields
 
Adding <math>1</math> on both sides yields
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As you can see,
 
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>
 
<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>
  
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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
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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>\sum_{i=0}^{2n} {10^i}=(\sum_{j=0}^n {(3 \times 10^j)})^2 + \sum_{k=0}^n {(2 \times 10^k)}</math>
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<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>
 
<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>
  
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Which proves PaperMath’s sum
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Which proves Papermath’s sum
 +
 
 +
==Problems==
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AMC 12A Problem 25
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 +
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>?
 +
 
 +
<math>\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20</math>
  
 
==Notes==
 
==Notes==
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Papermath’s sum was discovered by the aops user Papermath, as the name implies.
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PaperMath’s sum was discovered by the aops user PaperMath, as the name implies.
 
  
 
==See also==
 
==See also==
 +
*[[PaperMath’s circles]]
 
*[[Cyclic sum]]
 
*[[Cyclic sum]]
 
*[[Summation]]
 
*[[Summation]]
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[[Category:Algebra]]
 
[[Category:Algebra]]
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[[Category:Identities]]
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[[Category:Theorems]]

Revision as of 16:22, 27 March 2024

PaperMath’s sum

This is a summation identities for decomposition or reconstruction of summations. Papermath’s sum states,

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

Or

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

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

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

Proof

We will first prove a easier variant of Papermath’s sum,

$\sum_{i=0}^{2n} {(9 \times 10^i)}=(\sum_{j=0}^n {(9 \times 10^j)})^2 + 9\sum_{k=0}^n {(2 \times 10^k)}$

This is the exact same as

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

But everything is multiplied by $9$.

Notice that this is the exact same as saying

$\underbrace {9999\dots}_{2n}=(\underbrace {99\dots}_{n})^2+9(\underbrace {22\dots}_{n})$

Notice that $9(\underbrace {22\dots}_{n})=2(\underbrace {99\dots}_{n})$

Substituting this into $\underbrace {9999\dots}_{2n}=(\underbrace {99\dots}_{n})^2+9(\underbrace {22\dots}_{n})$ yields $\underbrace {9999\dots}_{2n}=(\underbrace {99\dots}_{n})^2+2(\underbrace {99\dots}_{n})$

Adding $1$ on both sides yields

$10^{2n}= (\underbrace {99\dots}_{n})^2+2(\underbrace {99\dots}_{n})+1$

Notice that $(\underbrace {99\dots}_{n})^2+2(\underbrace {99\dots}_{n})+1=(\underbrace {99\dots}_{n}+1)^2=(10^n)^2=10^{2n}$

As you can see,

$\sum_{i=0}^{2n} {(9 \times 10^i)}=(\sum_{j=0}^n {(9 \times 10^j)})^2 + 9\sum_{k=0}^n {(2 \times 10^k)}$

Is true since the RHS and LHS are equal

This equation holds true for any values of $n$. Since this is true, we can divide by $9$ on both sides to get

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

And then multiply both sides $x^2$ to get

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

Or

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

Which proves Papermath’s sum

Problems

AMC 12A Problem 25

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 $C_n - B_n = A_n^2$?

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

Notes

Papermath’s sum was discovered by the aops user Papermath, as the name implies.

See also

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