The Same Birthday
Although their ages are different, three children share the same birthday. On the last birthday, one of their ages was the sum of the other two. A few years later on their birthday, the youngest noticed that one of their ages was half the age of the other two added together. Since the birthday, on which one age was the sum of the other two, the number of years which had passed was half the sum of all the ages on the birthday when one age was the sum of the other two ... one was celebrating an 18th birthday ... which birthday were the other two celebrating?
(this is a test of Chris's english comprehension skills ... heh heh)
(this is a test of Chris's english comprehension skills ... heh heh)
posted by Zaux at
9:09 PM
18 Comments:
Hi guys ... I am burned out. I am goingg to let others post puzzles for a while. I need a break ... heh heh.
no kidding Zaux...take a break...you've earned it.this blog moved at 1 puzzle a day(if that),before you came along.rest up,take it easy,and after a while i hope you will give us one good puzzle a day(or whatever you feel like).
and...thank you.
Hear, hear. Zaux, you've been amazing.
Hear, hear. Zaux, you've been amazing.
English comprehension just caught Kgnirhmate's brain crap.
LOL
its 6, 12, and 18..
Sorry ... 6 and 12 is not correct
hi Zaux...are you sure?
6,12,18 looks like it works
18 years (and 1.21 gigawatts)later, they would be 24,30,36
(24+36)/2=30,right?
They're 14, 16, and 18.
When one age was the sum of two, it was 2, 4, 6.
When one age was half the sum of the other two, it was 10, 12, 14.
12 (2+4+6) years after the first set, they are 14, 16, and 18.
According to the puzzle source, the first birthday in question was 3,6, and 9. And the current one is 12, 15, and 18. If I get time later, I will take a look at it... or maybe by then, Chris or Cam will have a go at it.
This post has been removed by the author.
This post has been removed by the author.
Let a, b and c (c > b > a) be the ages when c = a+b.
Let y be the time that passed until one of them was 18.
Then we are given that y = (a+b+c)/2.
We are also given that (b+y) = ((a+y)+(c+y))/2 => b = (a+c)/2.
Substituting for c (from c=a+b) =>
b = a/2 + a/2 + b/2 => b = 2a and so c = a+b = 3a.
Using y = (a+b+c)/2 = (a+2a+3a)/2 = 3a
So when one of the kids is 18,
A is 4a, B is 5a and C is 6a years old. C must be the 18 year old, otherwise the others would have a fractional age, which isn't possible as they share the same birthday..
So 18 = 6a => a = 3, b = 6, c = 9 and y = 9.
So A is 12 and B is 15 and C is 18.
Which must be a great relief for Zaux ;)
Thanks Chris ... a definite relief
It took me much longer to decode the question than to do the 'rithmetic.
Hi Austin. Your (2+4+6) should have been divided by 2 => 6, which then gives 8, 10, 12 (but not 18).
Chris ...
I knew you would love the wording ...
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