Difference between revisions of "PaperMath’s sum"

Revision as of 20:48, 23 May 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.

@@ Line 25: / Line 25: @@
 <math>\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20</math>
-([[AMC 12A Problem 25|Source]])
+([[2018 AMC 10A Problems/Problem 25|Source]])
 == Notes ==

Page

Toolbox

Search