An algebra problem by David Altizio

Algebra Level 3

Let $p$ and $q$ be real numbers with $|p|<1$ and $|q|<1$ such that $p+pq+pq^2+pq^3+\cdots=2\qquad\text{and}\qquad q+qp+qp^2+qp^3+\cdots=3.$ What is $100pq$ ?

The answer is 48.

2 solutions

David Altizio
Apr 5, 2014

Note that by the formulae for geometric series the first equation can be rewritten as $\frac p{1-q}=2$ and the second can be rewritten as $\frac q{1-p}=3$ . This gives the system of equations $\begin{cases}p&=2-2q,\\q&=3-3p.\end{cases}$ Solving this gives $(p,q)=(\frac45,\frac35)$ , and so $100pq=100(\frac{12}{25})=\boxed{48}.$

I'm not really sure what why there are the conditions $|p | < 1$ and $|q| < 1$ . If either was greater than one, then at least one of the equations would diverge, giving no solutions. Therefore, even if you don't state $|p | < 1$ and $|q| < 1$ , the problem solver can figure it out for themselves .

Other than that, it's a fairly nice 'n simple problem. I like :)

Daniel Liu - 7 years, 2 months ago

Hmm, good point. I guess when I wrote this for my mock AMC a couple of months ago I wanted to avoid ambiguity at all costs.

David Altizio - 7 years, 2 months ago

good problem!

David Lee - 7 years, 2 months ago

The sum of an infinite GP is a/1-r only when r>1 which contradicts the question.???? Please explain.

Adarsh Kumar - 7 years, 2 months ago

Other way around. If r>1 then the series must diverge. (Just consider the sum 1+2+4+8+...)

David Altizio - 7 years, 2 months ago

hey then how can we solve it i.e. the example you give

Rishabh Jain - 7 years ago

I did the same way

Mas Mus - 7 years, 2 months ago

Prince Kumar
Apr 15, 2014

infinite geometric series n solve it by aljebra

p&q both are equal to 1. therefore 100 p q are equal to 37.5 . pl mail because am not satisfy with answer shown above.

amar nath - 7 years, 1 month ago

An algebra problem by David Altizio

The answer is 48.

2 solutions

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