Two friends meet after a long time in a bus. The first tells to the other that he has 3 kids. The product of their ages is 36. The sum of their ages is equal to the number of the bus they use. The second who has been a mathematician could not find the ages of the kids. So the father tell them that the biggest one has blue eyes.
How old are the kids?
2006-12-19 03:06:30 · 8 answers · asked by Job is done 1 in Science & Mathematics ➔ Mathematics
36 is divisible by 1, 2, 4, 6, 9, 18 and 36. There are several way to take 3 numbers that will multiply to 36, but only two ways to have numbers that will also add to the same total, that is 2, 2 and 9 (product 36, sum 13) and 1, 6 and 6 (product 36, sum also 13). Why the same total? Because the mathematician could not decide from the total, since he evidently knows he is riding bus #13. But he was told the oldest has blue eyes, so there is only one oldest kid (as opposed to 6 year old twins), so the oldest kid is 9 and there are 2 year old twins; so 9, 2 and 2 is the age of the kids.
2006-12-19 03:17:37 · answer #1 · answered by Vincent G 7 · 4⤊ 1⤋
The kids are 2, 2, and 9.
Three numbers that have a product of 36:
1 × 1 × 36 = 36
1 × 2 × 18 = 36
1 × 3 × 12 = 36
1 × 4 × 9 = 36
1 × 6 × 6 = 36
2 × 2 × 9 = 36
2 × 3 × 6 = 36
3 × 3 × 4 = 36
Sum of these ages:
1 + 1 + 36 = 38
1 + 2 + 18 = 21
1 + 3 + 12 = 16
1 + 4 + 9 = 14
1 + 6 + 6 = 13
2 + 2 + 9 = 13
2 + 3 + 6 = 11
3 + 3 + 4 = 10
They have to ride bus #13 because if they rode any other # bus, any of the other answers *could* be right and this problem would be ambiguous. The two sums that add up to 13 are 1, 6, and 6, and 2, 2, and 9.
Since there is an oldest one that has blue eyes, it has to be 2, 2, and 9.
2006-12-19 03:19:00 · answer #2 · answered by Ray H 3 · 1⤊ 1⤋
Clue 1: the product of the ages is 36. The factors of 36 are 1, 2, 2, 3 and 3. Therefore the ages may be 1, 4 and 9 (sum: 14); 1, 6 and 6 (sum: 13); 1, 3 and 12 (sum: 16); 1, 2 and 18 (sum: 21); 2, 2 and 9 (sum: 13); 2, 3 and 6 (sum: 11); or 3, 3 and 4 (sum: 10).
Clue 2: The sum of the ages is equal to the bus number AND the friend cannot, from this information, deduce the ages. He must know the bus number, so there must be two combinations of ages which both add up to the bus number. The only sum that corresponds to two age combinations is 13, so that must be the bus number, and the kids are either 1, 6 and 6 or 2, 2 and 9.
Clue 3: The biggest one has blue eyes. If the two oldest were twins, there wouldn't be a biggest one, so they can't be 1, 6 and 6 and therefore they must be 2, 2 and 9.
The kids are 2, 2, and 9 years old.
2006-12-19 03:18:39 · answer #3 · answered by Amy F 5 · 2⤊ 1⤋
Ok, we know that there are 3 kids, x*y*z = 36, and x+y+z=bus #. The color of the eyes tells us nothing about age of any of the kids, and the # of the bus is not known nor do we know how many digits are in the bus #. So; 1*6, 2*18, 3*12, 4*9, 6*6 =36, then since there are 3 kids the factorials break down to the possibilities {1,6,6} , {2,2,9} , {2,3,6} , {3,3,4}. Since the odds of twins is less than not I would choose {2,3,6}. Now that I posted mine and read the other answers, Why assume the sum must be 13? there is not enough information to get a absolute answer.
2006-12-19 04:39:00 · answer #4 · answered by Mike 1 · 0⤊ 1⤋
This is a bad, old question.
Marilyn vos Savant published it once in her column, "Ask Marilyn". I told her it was bad then, and I'll tell you why now.
Just because two kids have an (integral) age of 6, does NOT mean that they are twins!
It is certainly possible for a woman to give birth twice in one year. e.g., She gives birth in early January, gets pregnant again within 2.5 months, and gives birth again late that year. 6 years later, and before the first one's birthday, they are both 6. Therefore this is another case in which one can be an eldest (or in this case, biggest); so the problem is still indeterminant.
2006-12-19 03:43:06 · answer #5 · answered by ? 3 · 0⤊ 0⤋
Log off the computer and go find a RL human to talk to. The advice you get here will be a mixture of "go do someone to find out" and "repent sinner". Not stuff you need right now. Okay, so you're working out some heavy stuff pretty young. More power to you. You'll have a long road ahead, but you're going to be fine :D From the scenario you described (one I remember going through at your age) you're very likely a lesbian, hon. But it's no guarantee. Only you can figure that out. I wish you luck in your self searching.
2016-05-23 07:23:41 · answer #6 · answered by Anonymous · 0⤊ 0⤋
I asked that one yesterday...
2006-12-19 03:39:25 · answer #7 · answered by Anonymous · 0⤊ 1⤋
Kid #1: 9
Kid #2: 4
Kid #3:1
2006-12-19 03:18:39 · answer #8 · answered by Anonymous · 0⤊ 5⤋