Difference between revisions of "2014 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 3"
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== Solution == | == Solution == | ||
| + | The product ages is 72. The largest age is unique. The smallest age is unique. | ||
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| + | Assume that the youngest child is 1. Then the remaining 3 ages multiply to 72. The next smallest age possible is 2. The product of the final 2 ages is then 36 and this results in several solutions: | ||
| + | |||
| + | 1. (1,2,2,18) (can be disregarded) | ||
| + | |||
| + | 2. (1,2,3,12) | ||
| + | |||
| + | 3. (1,2,4,9) | ||
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| + | If the 2nd youngest is 3: | ||
| + | |||
| + | 1. (1,3,3,8) | ||
| + | |||
| + | 2. (1,3,4,6) | ||
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| + | If the 2nd youngest is 4, the final 2 ages multiply to 18. This leads to an impossibility. | ||
| + | |||
| + | Assume the youngest age is 2. Then the remaining ages multiply to 36. The next smallest age possible is 3: leads to a final solution of (2,3,3,4). | ||
| + | |||
| + | So we have 6 solutions. The first solution can be disregarded with knowledge of the age limit of AMC 8. | ||
== See also == | == See also == | ||
Latest revision as of 20:42, 25 July 2016
Problem
Two people, call them 
and 
, are having a discussion about the ages of ’s children.
A: “What are the ages, in years only, of your four children?”
B: “The product of their ages is 
.”
A: “Not enough information.”
B: “The sum of their ages equals your eldest daughter’s age.”
A: “Still not enough information.”
B: “My oldest child who is at least a year older than her siblings took the AMC 8 for the first time this year.”
A: “Still not enough information.”
B: “My youngest child is my only son.”
A: “Now I know their ages..”
What are their ages?
Solution
The product ages is 72. The largest age is unique. The smallest age is unique.
Assume that the youngest child is 1. Then the remaining 3 ages multiply to 72. The next smallest age possible is 2. The product of the final 2 ages is then 36 and this results in several solutions:
1. (1,2,2,18) (can be disregarded)
2. (1,2,3,12)
3. (1,2,4,9)
If the 2nd youngest is 3:
1. (1,3,3,8)
2. (1,3,4,6)
If the 2nd youngest is 4, the final 2 ages multiply to 18. This leads to an impossibility.
Assume the youngest age is 2. Then the remaining ages multiply to 36. The next smallest age possible is 3: leads to a final solution of (2,3,3,4).
So we have 6 solutions. The first solution can be disregarded with knowledge of the age limit of AMC 8.
See also
| 2014 UNM-PNM Contest II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
| All UNM-PNM Problems and Solutions | ||
