Difference between revisions of "PaperMath’s sum"

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==See also==
 
==See also==
*[[PaperMath’s circles]]
 
 
*[[Cyclic sum]]
 
*[[Cyclic sum]]
 
*[[Summation]]
 
*[[Summation]]

Revision as of 22:33, 1 September 2024

PaperMath’s sum

Papermath’s sum states,

$\sum_{i=0}^{2n-1} {(x^2 \times 10^i)}=(\sum_{j=0}^{n-1}{(3x \times 10^j)})^2 + \sum_{k=0}^{n-1} {(2x^2 \times 10^k)}$

Or

$x^2\sum_{i=0}^{2n-1} {10^i}=(x \sum_{j=0}^{n-1} {(3 \times 10^j)})^2 + x^2\sum_{k=0}^{n-1} {(2 \times 10^k)}$

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

$\sum_{i=0}^{2n-1} {10^i}=(\sum_{j=0}^{n -1}{(3 \times 10^j)})^2 + \sum_{k=0}^{n-1} {(2 \times 10^k)}$

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