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Devil's Dozen!!!

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Joined: 21 Aug 2017
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Re: Devil's Dozen!!!  [#permalink]

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New post 18 Apr 2018, 16:52
Bunuel wrote:
9. The product of three distinct positive integers is equal to the square of the largest of the three numbers, what is the product of the two smaller numbers?

700 question.

Let the three integers be \(a\), \(b\), and \(c\), where \(0<a<b<c\). Given: \(abc=c^2\) --> \(ab=c\). Question: \(ab=c=?\)

(1) The average (arithmetic mean) of the three numbers is 34/3 --> \(a+b+c=34\) --> \(a+b+ab=34\) --> \((a+1)(b+1)=35\). Now, since \(a\) and \(b\) are integers, then \(a+1=5\) and \(b+1=7\). \(a=4\) and \(b=6\) --> \(ab=24\). Sufficient. (Notice that \(a+1=1\) and \(b+1=35\) is not possible since in this case \(a=0\) and we are told that all integers are positive).

(2) The largest number of the three distinct numbers is 24. Directly give the value of c. Sufficient.

Answer: D.


How did you get from a+b+ab=34 to (a+1)(b+1)=35 ?
I do not understand where the extra +1 came from to get from 34 to 35

Thanks
Re: Devil's Dozen!!! &nbs [#permalink] 18 Apr 2018, 16:52

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