Logical Reasoning Puzzle With Solution - 2IIM

Mr.Crank: I have three kids now.

Mr. Sum: To think you were single last time we met. You definitely have been busy.

Mr. Product: I wish not to be a pedant. But I think your last statement should read “You were single and not yet a father the last time we met.”

Mr. Crank (inspired by the attention to detail): Perhaps we can create some kind of puzzle with their ages

Mr. Sum: Let us go for it. Will I be able to get their ages if you told us the sum of their ages?

Mr. Crank: No, of course not.

Mr. Product: Will I be able to get their ages if you told us the Product of their ages?

Mr. Crank (with a wry smile): Not now either. But the beauty is you might get very close if I told you both. But only very close.

Mr. Sum and Mr. Product: That is interesting. Give us some time.

Mr. Crank: Of course. If you get all the sets, dinner is on me.

Detailed solution to this puzzle follows:

There is no specific method that can be used to solve this puzzle; trial-and-error combined with intelligent inferences based on the conditions will lead you to the answer.

Now from what we understand from the very first sentence: it is clear that the friends are meeting after a decade. Since this condition is not qualified in any other statement in the puzzle, we can safely assume that “a decade” means EXACTLY a decade.

From the next few sentences we can understand that Mr. Crank has 3 kids and that he did not have any kids back when the 3 friends met exactly 10 years ago. So, now we know that the maximum age of any kid will be 9 years 11 months and some days; for all practical purposes this means that the maximum age is considered as 9. Also, 0 is not considered as a valid age so the minimum possible age is 1.

So the numbers we will work with are: {1,2,3,4,5,6,7,8,9}

Next we have

Mr. Sum: Let us go for it. Will I be able to get their ages if you told us the sum of their ages?

Mr. Crank: No, of course not.

This tells us that the sum of the ages is not 27 (9 + 9 + 9) or 3 (1+1+1) or some such simple number. If that were the case, the sum alone would be able to give us the answer.

Mr. Product: Will I be able to get their ages if you told us the Product of their ages?

Mr. Crank (with a wry smile): Not now either.

This tells us that the product is not 125 (5 * 5 * 5) or 1(1 * 1 *1) or 648(9 * 9 * 8) where the product alone gives away the answer.

Now, comes the most interesting part.

But the beauty is you might get very close if I told you both. But only very close.

This statements made by Mr. Crank about his kids reveals that we would not be able to find out the ages of the kids using the sum or the product of the ages. What does this mean?? Let us consider an example:

If I say that there are three people such that the sum of their ages is 4 and the product of the ages is 2 and none of them are older than 9 years, can you arrive at the ages? After some trial-and-error we can say that the ages would be (1, 1, 2); sum=4, product=2. If you probe further, you will realize that there is no other set of numbers (ages) that can satisfy the condition stated above. So, this is NOT the kind of sets that Mr. Crank is looking for.

So, effectively, we need to find sets of numbers that cannot uniquely have a product and a sum. In other words, if we are told sum = x, and product = y, more than one set of possibility should give us these results. We need to find x and y such that this condition is satisfied.

This is fairly difficult (and time consuming) to find, so let us try and eliminate some cases. We have to remember that there could be three types of solution sets: All three ages equal, 2 Ages equal and all three ages unique.

So, the number of cases will be :>

All Equal = 9C1 = 9; 2 Equal = 9C2 x 2C1 = 72; All three unique = 9C3 = 84. (This is just an interesting detour, not necessarily important for solving the sum.)

Let us try and eliminate some cases by looking at specific numbers.

• If all the ages are equal, then there cannot be another set of numbers (Within the limits of 1 and 9) with the same sum and product. This is easy to check as well since there are only 9 such cases.
• The numbers 1, 9, 8 cannot be the repeating numbers. Why? 9 and 8 it is slightly obvious because both represent very high powers, so we will definitely need those numbers in the second pair as well to make sure that the product is same; so not possible. With 1, you can see that since you are reducing a ‘1’ and factorizing the composite number to make sure the product remains the same, the sum will definitely reduce.
• Also, a (Prime, Prime, Prime) set will never work since there is no way we can split/factorize the numbers to get a different set
• No set can have the number 5, since we will need the number 5 to make the product end with 5/0 in the second set as well; so splitting the other 2 numbers in any manner will always give us a different sum.
• 7 cannot be part of the solution set as well since we will need 7 in the second set to get a product that is a factor of 7, so for a similar reason as that for the number 5, the second set will always give us a different sum.

Building on, we can make two more inferences that might make life easy.

1. Between the two triplets no number can be common. If we cannot have two triplets {a, a, b} and {a, a, c} – that much is obvious. But could we have triplets of the form {a, b, c} and {a, d, e}. If this were the case then we would have b + c = d + e and bc = de. This is impossible (AM and GM are equal only if all numbers are equal). So, the triplets have nothing in common.
2. We know we are dealing with {1, 2, 3, 4, 6, 8, 9}. Within this, we can either put 3 and 6 in different triplets and leave 9 out of it completely; Or, have 9 in one of the triplets in which case the other triplet will have to have {3, 6}, {6, 6} or {3, 3}.

So, we can have

1. {3, 6, a} and {9, b, c} as the triplet, or
2. {3, 3, a} and {9, b, c} as the triplet, or
3. {6, 6, a} and {9, b, c} as the triplet, or
4. No 9 at all

{8, 6, 3} and {9, 4, 4} works. So, does {6, 6, 1} and {2, 2, 9}.

The other template would be {3, a, b} and {6, c, d} – {6, 6, 2} and {8, 3, 3} works in this case.

We found three sets that satisfy all conditions:

(Notice that the product of all the solution sets are multiples of 36)

A really tough question, way tougher than what one can expect in the exam. Hope you had fun deciphering the clues and having a go at this one.

Best wishes for CAT.

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