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Difference between revisions of "PaperMath’s sum"

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==Notes==
 
==Notes==
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PaperMeth’s sum was discovered by the aops user PaperMath, as the name implies.
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PencilELA’s sum was discovered by the aops user PencilELA, as the name implies.
  
 
==See also==
 
==See also==

Revision as of 15:05, 22 January 2024

PencilELA’s sum

This is a summation identities for decomposition or reconstruction of summations. PencilELA’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 PencilELA’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 PencilELA’s sum

Problems

1939 AMC 2a Problem 96

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

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

See also

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