Difference between revisions of "Mock AIME 1 2007-2008 Problems/Problem 6"
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<center><math>\begin{align*}\sum_{r=1}^{\infty}\sum_{c=1}^{\infty} \left(\frac{1}{(2p)^r}\right)\left(\frac{1}{p^c}\right) &= \left(\sum_{r=1}^{\infty} \frac{1}{(2p)^r}\right)\left(\sum_{c=1}^{\infty} \frac{1}{p^c}\right)\\ | <center><math>\begin{align*}\sum_{r=1}^{\infty}\sum_{c=1}^{\infty} \left(\frac{1}{(2p)^r}\right)\left(\frac{1}{p^c}\right) &= \left(\sum_{r=1}^{\infty} \frac{1}{(2p)^r}\right)\left(\sum_{c=1}^{\infty} \frac{1}{p^c}\right)\\ | ||
&= \left(\frac{1}{1-\frac{1}{2p}}\right)\left(\frac{1}{1-\frac{1}{p}}\right)\\ | &= \left(\frac{1}{1-\frac{1}{2p}}\right)\left(\frac{1}{1-\frac{1}{p}}\right)\\ | ||
| − | &= \frac{ | + | &= \frac{2p^2}{(2p-1)(p-1)}\end{align*}</math></center> |
Taking the denominator with <math>p=2008</math> (indeed, the answer is independent of the value of <math>p</math>), we have <math>m+n \equiv 2008^2 + (2008-1)(2\cdot 2008 - 1) \equiv (-1)(-1) \equiv 1 \pmod{2008}</math> (or consider [[FOIL]]ing). The answer is <math>\boxed{001}</math>. | Taking the denominator with <math>p=2008</math> (indeed, the answer is independent of the value of <math>p</math>), we have <math>m+n \equiv 2008^2 + (2008-1)(2\cdot 2008 - 1) \equiv (-1)(-1) \equiv 1 \pmod{2008}</math> (or consider [[FOIL]]ing). The answer is <math>\boxed{001}</math>. | ||
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With less notation, the above solution is equivalent to considering the product of the geometric series <math>\left(1+\frac{1}{2 \cdot 2008} + \frac{1}{4 \cdot 2008^2} \cdots\right)\left(1 + \frac{1}{2008} + \frac{1}{2 \cdot 2008} \cdots \right)</math>. Note that when we expand this product, the terms cover all of the elements of the array. | With less notation, the above solution is equivalent to considering the product of the geometric series <math>\left(1+\frac{1}{2 \cdot 2008} + \frac{1}{4 \cdot 2008^2} \cdots\right)\left(1 + \frac{1}{2008} + \frac{1}{2 \cdot 2008} \cdots \right)</math>. Note that when we expand this product, the terms cover all of the elements of the array. | ||
| − | By the geometric series formula, the first series evaluates to be <math>\frac{1}{1 - \frac{1}{2 \cdot 2008}} = \frac{2 \cdot 2008}{2 \cdot 2008 - 1}</math>. The second series evaluates to be <math>\frac{1}{1 - \frac{1}{2008}} = \frac{2008}{2008 - 1}</math>. Their product is <math>\frac{2008 \cdot 4016}{(2008-1)(2\cdot 2008 - 1)}</math>, from which we find that <math>m+n</math> leaves a residue of <math>1</math> upon division by <math>2008</math>. | + | By the geometric series formula, the first series evaluates to be <math>\frac{1}{1 - \frac{1}{2 \cdot 2008}} = \frac{2 \cdot 2008}{2 \cdot 2008 - 1}</math>. The second series evaluates to be <math>\frac{1}{1 - \frac{1}{2008}} = \frac{2008}{2008 - 1}</math>. Their product is <math>\frac{2008 \cdot 4016}{(2008-1)(2\cdot 2008 - 1)}</math>, from which we find that <math>m+n</math> leaves a residue of <math>1</math> upon division by <math>2008</math>. |
== See also == | == See also == | ||
Revision as of 17:42, 10 July 2009
Problem 6
A 
-array is a structured, infinite, collection of numbers. For example, a 
-array is constructed as follows:
1 \qquad \frac 13\,\ \qquad \frac 19\,\ \qquad \frac 1{27} \qquad &\cdots\\ \frac 16 \qquad \frac 1{18}\,\ \qquad \frac{1}{54} \qquad &\cdots\\ \frac 1{36} \qquad \frac 1{108} \qquad &\cdots\\ \frac 1{216} \qquad &\cdots\\ &\ddots
\end{align*}$ (Error compiling LaTeX. Unknown error_msg)In general, the first entry of each row is 
times the first entry of the previous row. Then, each succeeding term in a row is times the previous term in the same row. If the sum of all the terms in a 
-array can be written in the form 
, where 
and 
are relatively prime positive integers, find the remainder when 
is divided by 
.
Solution
Note that the value in the 
th row and the 
th column is given by 
. We wish to evaluate the summation over all 
, and so the summation will be, using the formula for an infinite geometric series:
&= \left(\frac{1}{1-\frac{1}{2p}}\right)\left(\frac{1}{1-\frac{1}{p}}\right)\\
&= \frac{2p^2}{(2p-1)(p-1)}\end{align*}$ (Error compiling LaTeX. Unknown error_msg)Taking the denominator with 
(indeed, the answer is independent of the value of 
), we have 
(or consider FOILing). The answer is 
.
With less notation, the above solution is equivalent to considering the product of the geometric series 
. Note that when we expand this product, the terms cover all of the elements of the array.
By the geometric series formula, the first series evaluates to be 
. The second series evaluates to be 
. Their product is 
, from which we find that leaves a residue of 
upon division by .
See also
| Mock AIME 1 2007-2008 (Problems, Source) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
