Difference between revisions of "PaperMath’s sum"

m (PaperMath’s sum)
(PaperMath’s sum)
 
(5 intermediate revisions by one other user not shown)
Line 2: Line 2:
 
Papermath’s sum states,
 
Papermath’s sum states,
  
<math>\sum_{i=0}^{2n-1} {(x^2 \times 10^i)}=(\sum_{j=0}^{n-1}{(3x \times 10^j)})^2 + \sum_{k=0}^{n-1} {(2x^2 \times 10^k)}</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>
  
 
Or
 
Or
  
<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-1} {10^i}=(3x \sum_{j=0}^{n-1} {(10^j)})^2 + 2x^2\sum_{k=0}^{n-1} {(10^k)}</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
Line 14: Line 14:
 
==Proof==
 
==Proof==
  
First, note that the <math>x^2</math> part is trivial multiplication.
+
First, note that the <math>x^2</math> part is trivial multiplication, associativity, commutativity, and distributivity over addition,
 
 
===Simple Proof ===
 
  
 +
Observing that
 +
<math>\sum_{i=0}^{n-1} {10^i} =
 +
(10^{n}-1)/9</math>
 +
and
 
<math>(10^{2n}-1)/9 = 9((10^{n}-1)/9)^2 + 2(10^n -1)/9</math>
 
<math>(10^{2n}-1)/9 = 9((10^{n}-1)/9)^2 + 2(10^n -1)/9</math>
 
+
concludes the proof.
=== Papermath's proof ===
 
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>
 
 
 
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>
 
 
 
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>
 
 
 
And then multiply both sides <math>x^2</math> to get
 
 
 
<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>
 
 
 
Or
 
 
 
<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>
 
 
 
Which proves Papermath’s sum
 
  
 
==Problems==
 
==Problems==
Line 73: Line 31:
  
 
==Notes==
 
==Notes==
Papermath’s sum was named by the aops user Papermath. The name is not widely used.
+
Papermath’s sum was named by the aops user Papermath, after noticing it in a solution to an AMC 12 problem. The name is not widely used.
  
 
==See also==
 
==See also==
*[[PaperMath’s circles]]
 
 
*[[Cyclic sum]]
 
*[[Cyclic sum]]
 
*[[Summation]]
 
*[[Summation]]

Latest revision as of 17:07, 13 October 2024

PaperMath’s sum

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

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 named by the aops user Papermath, after noticing it in a solution to an AMC 12 problem. The name is not widely used.

See also