Sum Problem
In a group of 10 people, each one is asked to write sum of the ages of the other 9. The sums form the 9 element-set (82, 83, 84, 85, 87, 89, 90, 91, 92). Find the ages of the youngest and the oldest in the group and the ages of the two persons with the same age.
Labels: logic, mathemagic
posted by Rajesh Lal at
11:21 AM
9 Comments:
If there are 10 people, shouldn't there be 10 elements?
This post has been removed by the author.
This one was pretty difficult.
Evidently, CWolf didn't see that two kids are the same age so there are nine unique sums.
Kids ages are: 5,6,7,7,8,10,12,13,14,15
I'll let someone else figure out the oldest, youngest and which two are both 7.
Ok, I can name them.
George is the oldest, 15
and Henry is the youngest, 5
And the twins, Harry and Larry, are both 7.
With the provided sums the difference in age are set and one is a known duplicate
Express age of each as difference between youngest
x+0,x+1,x+2,x+3,x+5,x+7,x+8,x+9,x+10,x+c
where x+c represents the duplicate
and c must = 0,1,2,3,5,7,8,9 or 10
Using the youngest sum of 92 we see
92=9x+45+c or x=(47-c)/9
Most people stop telling ages in fractions after they reach 4, so
if we are stuck using integer values for x, only 2 fits for c, leaving x=5
for the set of 5,6,7,7,8,10,12,13,14,15
ANSWER:
youngest 5, oldest 15, two persons with same age 7
If, however, x may be non integer values then we can arbitrarily choose c from the set 0,1,2,3,5,7,8,9 or 10
and solve for x using x=(47-c)/9
e.g. for c =4 x=40/9 leaving the set
(4+4/9,5+4/9,6+4/9,7+4/9,9+4/9,
11+4/9,11+4/9,12+4/9,13+4/9,14+4/9)
ANSWER:
youngest,oldest,shared age: it depends on which is the duplicate sum
Cam
Very good Cam, I enjoyed your reasoning.
Thanks Ragknot
Note: There is a typo in my example for fractional ages. c=7 in the example, not 4. 4 is not a valid selection for c.
Cam
How did I do it?
See...
http://ragknot.blogspot.com/
the ages are 5 6 7 7 8 9 10 11 12 13 14 and 15
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