The Three Children of Mr. and Mrs. Jones
|Fun:|| (2.61) |
|Difficulty:|| (2.14) |
Mr. and Mrs. Jones have three children, all under the age of 20, but none with the same age.
Last year, the product of the three ages was 224.
This year the product of the three ages is 360
What are the children's ages now?
Answer:5, 8, and 9
The prime factorization of 224 is 2^5*7.
Using this, the possible sets of three ages of which the product is 224 are:
2, 7, 16
4, 7, 8
2, 8, 14
Now to find out which one of these sets was their ages a year ago, add 1 to each number and multiply:
The only set which multiplies to 360 is the 5, 8, and 9 set, meaning that these are the children's current ages.
Comments on this teaser
|Posted by stil||07/18/10|
|Shame on you for blowing off formulas, especially when some were used mentally.
On paper, the ages of the children last year were a, b, and c.
abc = 224 and involves partitioning of 2*2*2*2*2*7 into the three factors.
Adding 1 to each of last year's ages
abc + ab + ac + bc + a + b + c +1 =
224 + ab + ac + bc + a + b + c +1 =
225 + ab + ac + bc + a + b + c =
ab + ac + bc + a + b + c = 135.
Because there is a single odd factor, ab + ac + bc must be even, a + b + c must be odd, and the odd factor must be one of the latter three addends. Let it be a that = 7, then bc must = 2^5 = 32.
7b + 7c + 32 + 7 + b + c = 135.
8b + 8c = 96.
b + c = 12.
It is clear that the correct partition of five factors all 2's is (2*2)*(2*2*2).
Last year's ages must be 4, 7, and 8; this year's 5, 8, and 9.|
|Posted by stil||07/19/10|
|How old are "how old are they?" puzzles? I do wish the editors did a better job of screening out duplicates that vary only in the particulars. How many 'how old are these three people' puzzles have already been posted? How many of the 5000+ variations must be posted before "new" ones are declared duplication.
That said, they used to be one my favorite kinds of puzzle.|
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