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How would you solve this problem?

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How would you solve this problem?

by vongdn » Wed Sep 29, 2010 9:27 pm
How would you solve this problem?

5^21 x 4^11 = 2 x 10^n What is n?

This is how I did it, but I'm thinking there is a shorter more eloquent way that I am missing.

5^21 x 4^11 = 2 x 10^n
20^11 x 5^10 = 2 x 10^n
100^10 x 20 = 20 x 10^(n-1)
100^10 = 10^(n-1)
(10^2)^10 = 10^(n-1)
10^20 = 10^(n-1)

n-1 = 20 .... n = 21

Thanks
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by Rahul@gurome » Wed Sep 29, 2010 9:47 pm
Solution:
5^21 * 4^11 = 5^21 * 2^22 = 5^21 * 2^21 * 2 = 10^21 * 2.
So n = 21.
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by vongdn » Wed Sep 29, 2010 9:54 pm
Rahul@gurome wrote:Solution:
5^21 * 4^11 = 5^21 * 2^22 = 5^21 * 2^21 * 2 = 10^21 * 2.
So n = 21.
Wow, thanks, that is so much simpler than the way I solved it. What is the best way to quickly see a solution like that? It took me a while to figure it out. Definitely not in 2 mins.
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by Rahul@gurome » Wed Sep 29, 2010 11:07 pm
Wow, thanks, that is so much simpler than the way I solved it. What is the best way to quickly see a solution like that? It took me a while to figure it out. Definitely not in 2 mins.
How would you solve this problem?
The question has 2*(10^n). So obviously you will need pairs of 2's and 5's because they will combine to give 10.
Try to break up 5^21 * 4^11 so that you get equal powers of 2's and 5's with other factors.
Now 4^11 is 2^22. This has 22 2's.
Now 5^21 has 21 5's.
So when we pair up, we get 21 pairs of 2's and 5's and one 2 extra.
5^21 * 4^11 = 5^21 * 2^22 = 5^21 * 2^21 * 2 = 10^21 * 2.
So n = 21.
Rahul Lakhani
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On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
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