Mock AIME 1 2007-2008 Problems/Problem 6

Problem 6

A -array is a structured, infinite, collection of numbers. For example, a -array is constructed as follows:

$\begin{align*} 1 \qquad \frac 13\,\ \qquad \frac 19\,\ \qquad \frac 1{27} \qquad &\cdots\\ \frac 16 \qquad \frac 1{18}\,\ \qquad \frac{1}{54} \qquad &\cdots\\ \frac 1{36} \qquad \frac 1{108} \qquad &\cdots\\ \frac 1{216} \qquad &\cdots\\ &\ddots \end{align*}$ (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:

$\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)\\

&= \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 .

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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