The birthday coincidence

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Posted · Report post

I remarked to a friend how surprised I was to learn that as few as 23 randomly chosen people give better than even odds of a shared birthday. She agreed, saying three of her friends shared her birthday. The product of their ages is 2450 cubic years, she said, and their sum, remarkably, is twice your age [meaning mine.] It was a statement, to be sure, but I knew from the twinkle in her eye it was also a challenge.

I scribbled on the back of an envelope for a moment. Well? she smiled, got you stumped? In fact I could not answer. I'm afraid I need another clue. OK, she said, I am older than any of the three, and my age is equal to the product of the ages of the two youngest.

Triumphantly I announced the ages of her three friends.

p.s. That's a challenge. ;)

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Posted · Report post

How about this:

Her friends are 2, 25, and 49

She is 50

You are 38

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Posted · Report post

How about this:

Her friends are 2, 25, and 49

She is 50

You are 38

If that was the case, why would bonanova need another clue?

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Posted · Report post

Factors of 2450 = 5x5x2x7x7


so their ages could be 5,10,49 or 25,2,49, or 5,7,70 or 7,7,50; these
are the factors out of the several, which match the logic of the OP.
Now Bonanova's friend could not be 14 or 15yr.. I hope. So 25,2,49 seems to be correct one.
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Posted · Report post

I don't think there are any integer solutions to this.

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Posted · Report post

Are we using your age as shown on your profile page (72)?

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Posted · Report post

Are we using your age as shown on your profile page (72)?

No, Sherlock, ;) but great investigative reporting!

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Posted · Report post

Are we using your age as shown on your profile page (72)?

No, Sherlock, ;) but great investigative reporting!

Oh that makes things different. In that case I think there are several solutions.

Actually, what was the point in saying that it is twice your age? All it tells us is that the sum is even, but we already know this as any 3 numbers formed from the prime factorisation of 2450 sum to be even.

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Posted · Report post

Are we using your age as shown on your profile page (72)?

No, Sherlock, ;) but great investigative reporting!
Oh that makes things different. In that case I think there are several solutions.

Actually, what was the point in saying that it is twice your age? All it tells us is that the sum is even, but we already know this as any 3 numbers formed from the prime factorisation of 2450 sum to be even.

Your statement is correct.

Your question is valid.

But since you don't yet have a unique answer, it's your question to answer.

Btw, no offense intended by the Sherlock comment. :)

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Posted · Report post

Are we using your age as shown on your profile page (72)?

No, Sherlock, ;) but great investigative reporting!
Oh that makes things different. In that case I think there are several solutions.

Actually, what was the point in saying that it is twice your age? All it tells us is that the sum is even, but we already know this as any 3 numbers formed from the prime factorisation of 2450 sum to be even.

Your statement is correct.

Your question is valid.

But since you don't yet have a unique answer, it's your question to answer.

Btw, no offense intended by the Sherlock comment. :)

None taken.

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Posted · Report post

Time's up.

The ages are 5, 10, and 49.


Bonanova is 32.
That is the only combination that does not result in a unique sum.
7+7+50 is also 64, but 7*7 < 50, whereas 5*10 > 49.

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