Difference between revisions of "Mock AIME 1 2007-2008 Problems/Problem 6"
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== Solution == | == Solution == | ||
Note that the value in the <math>r</math>th row and the <math>c</math>th column is given by <math>\left(\frac{1}{(2p)^r}\right)\left(\frac{1}{p^c}\right)</math>. We wish to evaluate the summation over all <math>r,c</math>, and so the summation will be, using the formula for an infinite [[geometric series]]: | Note that the value in the <math>r</math>th row and the <math>c</math>th column is given by <math>\left(\frac{1}{(2p)^r}\right)\left(\frac{1}{p^c}\right)</math>. We wish to evaluate the summation over all <math>r,c</math>, and so the summation will be, using the formula for an infinite [[geometric series]]: | ||
| − | < | + | <cmath>\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{2p^2}{(2p-1)(p-1)}\end{align*}</ | + | &= \frac{2p^2}{(2p-1)(p-1)}\end{align*}</cmath> |
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>. | ||
Revision as of 18:15, 29 January 2020
Problem 6
A 
-array is a structured, infinite, collection of numbers. For example, a 
-array is constructed as follows:
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:
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 | ||
