Difference between revisions of "PaperMath’s sum"
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− | == | + | == PaperMath’s sum== |
− | 1+ | + | This is a summation identities for decomposition or reconstruction of summations. 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> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | 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> | ||
==Proof== | ==Proof== | ||
− | We will first prove a easier variant of | + | 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> | <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> | ||
− | This is the exact same as | + | This is the exact same as |
<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} {10^i}=(\sum_{j=0}^n {(3 \times 10^j)})^2 + \sum_{k=0}^n {(2 \times 10^k)}</math> | ||
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Notice that <math>9(\underbrace {22\dots}_{n})=2(\underbrace {99\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 | + | 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> |
− | <math>\underbrace {9999\dots}_{2n}=(\underbrace {99\dots}_{n})^2+2(\underbrace {99\dots}_{n})</math> | ||
Adding <math>1</math> on both sides yields | Adding <math>1</math> on both sides yields | ||
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As you can see, | 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> | <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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<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} {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 | + | Which proves Papermath’s sum |
==Problems== | ==Problems== | ||
− | + | AMC 12A Problem 25 | |
For a positive integer <math>n</math> and nonzero digits <math>a</math>, <math>b</math>, and <math>c</math>, let <math>A_n</math> be the <math>n</math>-digit integer each of whose digits is equal to <math>a</math>; let <math>B_n</math> be the <math>n</math>-digit integer each of whose digits is equal to <math>b</math>, and let <math>C_n</math> be the <math>2n</math>-digit (not <math>n</math>-digit) integer each of whose digits is equal to <math>c</math>. What is the greatest possible value of <math>a + b + c</math> for which there are at least two values of <math>n</math> such that <math>C_n - B_n = A_n^2</math>? | For a positive integer <math>n</math> and nonzero digits <math>a</math>, <math>b</math>, and <math>c</math>, let <math>A_n</math> be the <math>n</math>-digit integer each of whose digits is equal to <math>a</math>; let <math>B_n</math> be the <math>n</math>-digit integer each of whose digits is equal to <math>b</math>, and let <math>C_n</math> be the <math>2n</math>-digit (not <math>n</math>-digit) integer each of whose digits is equal to <math>c</math>. What is the greatest possible value of <math>a + b + c</math> for which there are at least two values of <math>n</math> such that <math>C_n - B_n = A_n^2</math>? | ||
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==Notes== | ==Notes== | ||
− | + | Papermath’s sum was discovered by the aops user Papermath, as the name implies. | |
− | |||
==See also== | ==See also== | ||
+ | *[[PaperMath’s circles]] | ||
*[[Cyclic sum]] | *[[Cyclic sum]] | ||
*[[Summation]] | *[[Summation]] | ||
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[[Category:Algebra]] | [[Category:Algebra]] | ||
− | [[Category: | + | [[Category:Theorems]] |
Latest revision as of 17:22, 27 March 2024
Contents
PaperMath’s sum
This is a summation identities for decomposition or reconstruction of summations. Papermath’s sum states,
Or
For all real values of , this equation holds true for all nonnegative values of
. When
, this reduces to
Proof
We will first prove a easier variant of Papermath’s sum,
This is the exact same as
But everything is multiplied by .
Notice that this is the exact same as saying
Notice that
Substituting this into yields
Adding on both sides yields
Notice that
As you can see,
Is true since the RHS and LHS are equal
This equation holds true for any values of . Since this is true, we can divide by
on both sides to get
And then multiply both sides to get
Or
Which proves Papermath’s sum
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 discovered by the aops user Papermath, as the name implies.