The question is from CAT Ratio, Mixtures and Averages. Given the product of n digits, our task is to minimize the average of those numbers which is related to the sum of those numbers. CAT exam is known to test on basics rather than high funda ideas. A range of CAT questions can be asked from Ratios and Proportions, Mixtures, Alligations and Averages. Make sure you master the topics. 2IIMs CAT questions bank provides you with CAT questions that can help you gear for CAT Exam CAT 2020.

Question 47: If the product of n distinct positive integers is n^{n}. What is the minimum value of their average if n = 6?

- 6
- 10
- \\frac{46}{6}\\)
- 8

\\frac{46}{6}\\)

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For the average to be minimum, the numbers should be as close to each other as possible.

Why? Let us take a simpler example to check this out. Let us say product of two natural numbers is 64. We could take the two extremes 1 * 64 and 8 * 8. In the first instance the average is 32.5, in the second one the average is 8. The closer the numbers are, the smaller their sum and average are going to be.

In our case, we are looking at 6^{6}. If the numbers need not have been distinct, then 6 * 6 * 6 * 6 * 6 * 6 would work best. Now, let us anchor around 6 and try to figure out a mechanism of keeping numbers as close to 6 as possible.

We need to look for

1. 6 numbers multiplying to give 2^{6} * 3^{6}

2. With the numbers as close to 6 as possible

Let us get in a 6 obviously, then include 8 and 9 as well. 6 * 8 * 9 = 2^{4} * 3^{3}

So, the other 3 numbers should multiply to give 2^{2} * 3^{3}. There are more 3’s than 2’s, so perhaps we could include 18. So, we could have 6, 8, 9, 18 – these 4 multiply to give 2^{5} * 3^{5}. The last 2 could be 2 and 3.

So, we could have 2, 3, 6, 8, 9, 18. These numbers add up to 46 giving us an average of 46/6.

Between 2 and 18 the only other numbers that are factors of 66 that we have missed out on are 4 and 12. We will need to see if we can have some combination that includes either one or both of these numbers that can bring down the average.

Scenario I – Including 12.

Let us start by including 6 and 12. We have accounted for 2^{3} * 3^{2}. We have more 2s than 3’s in the system so we can perhaps offset this by including a 9.

If we had 6, 9 and 12, the product would be 2^{3} * 3^{4}. The remaining 3 numbers should give us 2^{3} * 3^{2}. We have already included 6, 9 and 12. So, if we need more multiples of 3, we need either 3 or 18. Merely having 3 does not help as we still need to accommodate 2 more 3’s. If we include 18, our average will go higher anyway. We will have 1, 4, 6, 9, 12, 18.

Let us rejig our approach. We need six 3’s. We can include 3, 6, 9 and 12 but that gets us to only 5 3’s. So, we need to have 18 or 24. 24 is too large and would mangle the average so we NEED to have an 18 in the mix. From 3, 6, 9, 12 we need to find 4 more 3’s. We cannot drop 9 from the mix as the other 3 would give us only three more 3’s. So, we will DEFINITELY need to have 9 also.

So, we have 9 and 18 in the mix. And 2 of 3, 6 and 12 in the mix. We could have

3, 6, 9, 18 , or 3, 9, 12, 18 or 6, 9, 12, 18 –

3, 6, 9, 18 – Product of remaining two numbers should be 24. Best case scenario would be 3, 6, 9, 18, 2, 8 yielding a sum of 46

3, 9, 12, 18 - – Product of remaining two numbers should be 23. Best case scenario would be 3, 9, 12, 18, 2, 4 yielding a sum of 48.

6, 9, 12, 18 – – Product of remaining two numbers should be 22. Best case scenario would be 6, 9, 12, 18, 1, 4 yielding a sum of 50.

Overall best-case scenario is 3, 6, 9, 18, 2, 8 yielding a sum of 46 and an average of 46/6

The question is **"What is the minimum value of their average if n = 6?"**

Choice C is the correct answer.

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