I am posting a problem after a long time. Try this out and post your answer in the comment section.
QUESTION:
A mother has four children, each with a di erent age. The product of their ages is 17 280. The sum of the ages of the three oldest children is 40 and the sum of the ages of the three youngest children is 32. Determine the ages of the four children.
Let the ages of the children from youngest to oldest be a; b; c; d.
ReplyDeleteSince the ages of the three oldest children sum to 40, b + c + d = 40. (1)
Since the ages of the three youngest children sum to 32, a + b + c = 32. (2)
Subtracting (2) from (1), we obtain d a = 8. This means that the di erence between the age
of the oldest child and the age of the youngest child is 8.
Now 17 280 = 2
7 3
3 5 = (2
2 3) (2
2 3) (2
2 3) 2 5 = 12 12 12 10. Since all
of the ages are di erent, this statement tells us that at least one child is over 12 and one child
is under 10.
There is a limited number of possibilities such that the di erence between the oldest and
youngest is 8 which also satisfy the condition that the youngest is under 10 and the oldest is
over 12. The possibilities for youngest and oldest are (5,13), (6,14), (7,15), (8,16), and (9,17).
No other combination would be possible since the oldest child must be over 12 and the
youngest child must be under 10.
The numbers 7, 13, and 17 are primes and are not factors of 17 280. Therefore we can
eliminate the possibilities where an age is one of 7, 13, or 17, leaving (6,14) and (8,16). But 14
contains a prime factor of 7 which is not a factor of 17 280. Since there is only one possibility
left, (8,16), we can conclude that the youngest is 8 and the oldest is 16.
Now 17 280 = 8 16 3
3 5. Using the remaining factors 3
3
and 5, we need to create two
numbers between 8 and 16. The only possibilities are 3
2 = 9 and 3 5 = 15.
Therefore the ages of the children are 8, 9, 15, and 16. It is easy to verify that this is the
correct solution.
Posted By Prakhar Bindal (Class 8) Meerut Dewan Public School
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