Hi,

I have the following trick problem with the solution, but, i don't understand the solution. Here is the problem and the solution:

My question is: How can i guess the following phrase?

What math rule am i missing? Can anyone explain me the solution?

Regards,
Pedro

Agreed, the wording on that phrase in the solution is kind of awkward.

Let's think about this: What are all the multiples of 36?
1, 2, 3, 4, 6, 9, 12, 13 and 36.
This means that there are 4 solutions (that I can come up with, on the spot) where she has three children with an oldest child:
Sol'n 1: 1 x 2 x 18
Sol'n 2: 2 x 3 x 6
Sol'n 3: 2 x 2 x 9
Sol'n 4: 3 x 3 x 4

Where the book's solution gets the idea about 14 is really confusing me. I bet if you want more information on the author's rationalization on the answer, you would need to know the address of the house next door.



Here's another one:
What's the percent chance you will roll a 6 (six) with one die?

Statistical Sol'n: ~16.7%
Realistic Sol'n: 50% - You either roll it, or you don't.

Yeah, it is an obstinate answer, but it drove my evil Data Management teacher up the wall, and that's what counts.

I think you mean ~16.7%.

not if it's a special die that has 600 sides, and only one of which has a six on it. :-D

The multiples are 1:36 2:18 3:12 4:9 6:6. Now, of course you need to further break these down to determine which sets are likely to cause an intersection with sums. In order the possible terms are: 1, 2, 3, 4, 6, 9, 12, 18, 36...

And I think all the three term solutions are:
1 x 1 x 36 = 36 (sum=38)
1 x 2 x 18 = 36 (sum=21)
1 x 3 x 12 = 36 (sum=16)
1 x 4 x 9 = 36 (sum=14)
1 x 6 x 6 = 36 (sum=13)
2 x 2 x 9 = 36 (sum=13)
2 x 3 x 6 = 36 (sum=11)
3 x 3 x 4 = 36 (sum=10)

Note: We handle the terms in order... so that we avoid repeats. If we need to use a number before the current one to reach 36 in 3 steps (like 4 x 9 x 1 we know we already have an equivalent form before that... 1 x 4 x 9 in this case).

As you can see... only these two have the same sum. And one of them fails the "oldest child" test. By my count, there are 7 possible situations where there is an oldest child. All except the 1 x 6 x 6.

Edit to change it to 7 in the line right above this one, not 8 like I originally counted. Would you believe that counting the number of items was harder for me than doing the stuff above it? What the hell is wrong with me?!

Why does the sum of their ages have to equal 13? The only facts that have been presented in the original quote are that there are 3 children; the product of their ages is 36; and that the sum of their ages equals the number of the house next door. Nowhere is it stated that the sum must be 13, or that the house next door is #13. Therefore, and in the absence of any disclosure as to what the next house's number actually is, any combination of numbers that equal 36 when multiplied together is equally likely. In other words, if the house were #14 or #38, etc, then any of the combinations frob23's listed above are equally valid. Along these lines, if we consider that the woman's children could involve twins or triplets (*one* kid would be the eldest no matter what) then you could have two kids of the same age, which complicates things further.

Were there conditions that were omitted from the original problem description? Providing critical, relevant info only in the 'solution' makes the problem impossible to solve.....

What we know, from the problem.
---
Their ages multiplied equal 36.
The sum of their ages is the house number next door.
The census taker had to come back after seeing the number next door.

The reason we know it had to be 13 is that the census taker had to come back. If the house next door had been any other number, he would have known the ages of the children. The only way he could have needed more information was if the sum itself didn't define a discrete set of numbers. We know she doesn't have triplets, simply because 36 doesn't have a simple cube root. If it was 27 or 64 then we would have to deal with the possibility of triplets.

The solution itself requires that she have twins. And that the twins are either 6 or 2. So we do have to deal with the "eldest twin" possibility. But, in practice, a parent doesn't refer to the first twin as being elder to the other in common conversation. As much as it drives the twins crazy, they are usually referred to as "the twins." My mother is an identical twin and this still happens when people refer to her and her sister when they were little. I know other sets of twins (simply because my mother[s] love meeting others and people with twins) and the reference to them by birth order is virtually unknown -- mostly because it isn't a way to distinguish between them. When we say "oldest" or "youngest" in common speech we expect that, when presented by the people being talked about, that information will be sufficient for us to know who is who.

So, this question does hinge on a linguistic understanding of how "oldest child" would be used in a situation. Still, we can be sure the ages had to be of the two sets offered in the above solution and marked bold in my response, simply because there was a confusion by the census taker. And then, determining which set is the correct one requires knowing the little bit about verbal references to twins.