Difference between revisions of "PaperMath’s sum"

Latest revision as of 23:14, 9 November 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 due to its randomness.

@@ Line 1: / Line 1: @@
 == PaperMath’s sum==
-This is a summation identities for decomposition or reconstruction of summations. Papermath’s sum states,
+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-1} {(10^ix^2)}=(\sum_{j=0}^{n-1}{(10^j3x)})^2 + \sum_{k=0}^{n-1} {(10^k2x^2)}</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>
+<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
-<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-1} {10^i}=(\sum_{j=0}^{n -1}{(3 \times 10^j)})^2 + \sum_{k=0}^{n-1} {(2 \times 10^k)}</math>
 ==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>
+First, note that the <math>x^2</math> part is trivial multiplication, associativity, commutativity, and distributivity over addition,
-This is the exact same as
+Observing that
+<math>\sum_{i=0}^{n-1} {10^i} =
-<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>
+(10^{n}-1)/9</math>
+and
-But everything is multiplied by <math>9</math>.
+<math>(10^{2n}-1)/9 = 9((10^{n}-1)/9)^2 + 2(10^n -1)/9</math>
+concludes the proof.
-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==
@@ Line 65: / Line 31: @@
 ==Notes==
-Papermath’s sum was discovered by the aops user Papermath, as the name implies.
+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 due to its randomness.
 ==See also==
-*[[PaperMath’s circles]]
 *[[Cyclic sum]]
 *[[Summation]]

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