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| ==Proof== | | ==Proof== |
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− | 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 ===
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| + | Observing that |
| <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 ===
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− | We will first prove a easier variant of Papermath’s sum,
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− | <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>
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− | This is the exact same as
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− | <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>
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− | But everything is multiplied by <math>9</math>.
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− | Notice that this is the exact same as saying
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− | <math>\underbrace {9999\dots}_{2n}=(\underbrace {99\dots}_{n})^2+9(\underbrace {22\dots}_{n})</math>
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− | Notice that <math>9(\underbrace {22\dots}_{n})=2(\underbrace {99\dots}_{n})</math>
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− | 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>
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− | Adding <math>1</math> on both sides yields
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− | <math>10^{2n}= (\underbrace {99\dots}_{n})^2+2(\underbrace {99\dots}_{n})+1</math>
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− | 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>
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− | As you can see,
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− | <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>
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− | Is true since the RHS and LHS are equal
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− | 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
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− | <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>
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− | And then multiply both sides <math>x^2</math> to get
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− | <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>
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− | Or
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− | <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>
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− | Which proves Papermath’s sum
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| ==Problems== | | ==Problems== |
Revision as of 21:30, 1 September 2024
PaperMath’s sum
Papermath’s sum states,
Or
For all real values of , this equation holds true for all nonnegative values of . When , this reduces to
Proof
First, note that the part is trivial multiplication, associativity, commutativity, and distributivity over addition,
Observing that
concludes the proof.
Problems
AMC 12A Problem 25
For a positive integer and nonzero digits , , and , let be the -digit integer each of whose digits is equal to ; let be the -digit integer each of whose digits is equal to , and let be the -digit (not -digit) integer each of whose digits is equal to . What is the greatest possible value of for which there are at least two values of such that ?
Notes
Papermath’s sum was named by the aops user Papermath. The name is not widely used.
See also