If the product of 6 distinct positive integers is 6^6. What is the min

Hint : Numbers are 1, 2, 3 , 12 , 18, 36

average is 12 for the above numbers

Taking 1 as one of th numbers has no merit because it otherwise makes the bigger number even more bigger.

Hi, is the concept of GM tested on GMAT? I haven't studied it yet.. if so, will go through some material!

For a given product of n numbers, the sum of the numbers (and hence, the average) is minimized if the terms are all equal.

For example: If we have 2 terms, having product 9, the minimum sum is 6 when each is 3 (other cases are 1 and 9 => sum is 10, greater than 6 which happens when each is 3)

Again, for example: If we have 3 terms, having product 64, the minimum sum is 12 when each is 4 (other cases are 1, 1 and 64 => sum is 66; Or 2, 4 and 8 => sum is 14; etc., greater than 12)

This is represented as: For a given set of numbers, the Arithmetic Mean is ≥ Geometric Mean


For our question: For the given product as 6^6:
To minimize the average, we need to minimize the sum => Each number must be 6

=> Average = 6

Answer B

sujoykrdatta

Great theory explanation.

But for the given question, aren't the numbers supposed to be distinct?

Yes, you are right. Fixed. Thank you.

Any better method to do this , think it’s not possible to solve in 2 min

Posted from my mobile device

Yes, this is a bit tricky and I would do it considering the concept of deviations.

If distinct numbers were not necessary, I would make every number 6 and get the mean 6. This would be the smallest possible mean.
But distinct numbers are required. So I will try to limit the deviations around 6. I would try to keep the deficit and excess as balanced as possible so that mean doesn't change much.

First few factors of 2^6 * 3^6 are:

1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, ....

3s will increase the number very quickly and I need to consume six 3s. So I will try to get the efficient numbers (the small numbers that use up 3s). Also, I know the excess will increase rapidly the moment I pick big numbers but deficit will not so I will try to keep as many small numbers utilising 3s as possible.

So I pick 3, 6, 9, (four 3s are chosen). Next is 12 but it uses only one 3. So I will instead pick 18 which uses up the other two 3s too and is not very large. Otherwise, had we picked 12, we would have needed to pick 24 too which is much greater.


Selected Numbers - 3, 6, 9, 18
All 3s are consumed. But they use only two 2s yet (in 6 and 18). So we need two more numbers with four 2s. Now it is easy to pick 2 and 8.

So selected numbers - 2, 3, 6, 8, 9, 18
Average = 23/3

Answer (C)

Deviation concept is quite useful. Check out this blog post for another discussion on it:
https://anaprep.com/arithmetic-usefulne ... eviations/

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