Logarithms in 2020

Algebra Level 2

log a x + log a y = log a 2020 \log_{a}{x} + \log_{a}{y} = \log_{a}{2020}

For positive rational a a , how many ordered pairs of positive integers ( x , y ) (x,y) are possible?


This problem is not original


The answer is 12.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Mahdi Raza
Apr 14, 2020

log a x + log a y = log a 2020 log a ( x y ) = log a 2020 x y = 2020 2020 = 2 2 5 1 10 1 1 Factors of 2020 = ( 2 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 12 \begin{aligned} \log_{a}{x} + \log_{a}{y} &= \log_{a}{2020} \\ \log_{a}{(x\cdot y)} &= \log_{a}{2020} \\ xy &= 2020 \\ \\ 2020 &= 2^{2} \cdot 5^{1} \cdot 101^{1} \\ \text{Factors of 2020} &= (2+1)(1+1)(1+1) \\ &= \boxed{12} \end{aligned}

Since each factor corresponds to another factor to multiply to 2020, there are 12 pairs in total \text{Since each factor corresponds to another factor to multiply to 2020, there are 12 pairs in total}

x y 1. 1 2020 2. 2 1010 3. 4 505 4. 5 404 5. 10 202 6. 20 101 7. 101 20 8. 202 10 9. 404 5 10. 505 4 11. 1010 2 12. 2020 1 \begin{array}{c|c|c} &x&y \\ \hline \\ 1.&1&2020 \\ 2.&2&1010 \\ 3.&4&505 \\ 4.&5&404 \\ 5.&10&202 \\ 6.&20&101 \\ 7.&101&20 \\ 8.&202&10 \\ 9.&404&5 \\ 10.&505&4 \\ 11.&1010&2 \\ 12.&2020&1 \end{array}

April 14 problem in "new" section?!

Vinayak Srivastava - 1 year ago

Log in to reply

I don't know why

Mahdi Raza - 1 year ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...