Ages of Three Children puzzle: Difference between revisions
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== Solutions == |
== Solutions == |
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=== for |
=== for 72 === |
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The [[Integer factorization|prime factors]] of 72 are 2, 2, 2, 3, 3; in other words, {{math|2 × 2 × 2 × 3 × 3 {{=}} 72}} |
The [[Integer factorization|prime factors]] of 72 are 2, 2, 2, 3, 3; in other words, {{math|2 × 2 × 2 × 3 × 3 {{=}} 72}} |
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Revision as of 15:49, 17 June 2020
The Ages of Three Children puzzle is a logic puzzle which on first inspection seems to have insufficient information to solve, but which rewards those who persist and examine the puzzle critically.
The puzzle
A census taker approaches a woman leaning on her gate and asks about her children. She says, "I have three children and the product of their ages is seventy–two. The sum of their ages is the number on this gate." The census taker does some calculation and claims not to have enough information. The woman enters her house, but before slamming the door tells the census taker, "I have to see to my eldest child who is in bed with measles." The census taker departs, satisfied.[1]
The problem can be presented in different ways, giving the same basic information: the product, that the sum is known, and that there is an oldest child (e.g. their ages adding up to today's date, [2] or the eldest being good at chess[3]).
Another version of the puzzle gives the age product as thirty–six, which leads to a different set of ages for the children.[4]
Solutions
for 72
The prime factors of 72 are 2, 2, 2, 3, 3; in other words, 2 × 2 × 2 × 3 × 3 = 72
This gives the following triplets of possible solutions;
| Age one | Age two | Age three | Total (Sum) |
|---|---|---|---|
| 1 | 1 | 72 | 74 |
| 1 | 2 | 36 | 39 |
| 1 | 3 | 24 | 28 |
| 1 | 4 | 18 | 23 |
| 1 | 6 | 12 | 19 |
| 1 | 8 | 9 | 18 |
| 2 | 2 | 18 | 22 |
| 2 | 3 | 12 | 17 |
| 2 | 4 | 9 | 15 |
| 2 | 6 | 6 | 14 |
| 3 | 3 | 8 | 14 |
| 3 | 4 | 6 | 13 |
Because the census taker knew the total (from the number on the gate) but said that he had insufficient information to give a definitive answer; thus, there must be more than one solution with the same total.
Only two sets of possible ages add up to the same totals:
A. 2 + 6 + 6 = 14
B. 3 + 3 + 8 = 14
In case 'A', there is no 'eldest child' - two children are aged six (although one could be a few minutes or around 9 to 12 months older and they still both be 6). Therefore, when told that one child is the eldest, the census-taker concludes that the correct solution is 'B'.[2]
for 36
The prime factors of 36 are 2, 2, 3, 3 This gives the following triplets of possible solutions;
| Age one | Age two | Age three | Total (Sum) |
|---|---|---|---|
| 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 |
Using the same argument as before it becomes clear that the number on the gate is 13, and the ages 9, 2 and 2.[4]
References
- ^ "Ask Dr. Math". Math Forum. 2008-11-22. Archived from the original on 30 August 2010. Retrieved 2010-09-12.
- ^ a b Mary Jane Sterling (2007), Math Word Problems for Dummies, For Dummies, p. 209, ISBN 978-0-470-14660-6, retrieved 2010-09-12
- ^ Rick Billstein; Shlomo Libeskind; Johnny W. Lott (1997), A problem solving approach to mathematics for elementary school teachers (6 ed.), Addison-Wesley, ISBN 978-0-201-56649-9
- ^ a b "Math Puzzle - Census - Solution". Mathsisfun.com. Archived from the original on 3 September 2010. Retrieved 2010-09-12.