# Enigmatic Code

Programming Enigma Puzzles

## Puzzle #61: Triple jump

From New Scientist #3284, 30th May 2020 [link] [link]

“My Auntie Connie just had triplets, three boys!”
“Wow, how old is she?”
“I dunno, but she’s really old.”
“Do you think the boys’ ages will ever catch up to your auntie’s? If you add all three together, I mean.”
“I’m not sure they’ll ever add up to her age exactly — I think it depends on how old she is now.”

Assuming they all live to a ripe old age, what are the chances that there will come a date in the future when the ages of the three boys add up exactly to their mother’s age?

(To be clear: your “age” is how old you were on your last birthday, so it is always a whole number).

[puzzle#61]

### 3 responses to “Puzzle #61: Triple jump”

1. Jim Randell 30 May 2020 at 8:14 am

I assume we are dealing with ages in whole numbers of years. And the triplets celebrate their birthdays on the same day.

Suppose the mother’s age was n when the triplets were born.

Then after k years have passed the cumulative age of the triplets is 3k and the age of the mother is (n + k).

These have the same value when:

3k = n + k
2k = n
k = n / 2

So if the mother’s age is an even number when the triplets were born, there will be a time when their cumulative age matches their mother’s current age.

For example, if the mother was 30 when the triplets were born, then after 15 years the triplets would have a cumulative age of 45, and the mother’s current age would also be 45.

But if she was 31, then after 15 years the triplets would have a cumulative age of 45, and the mother’s age would be 46. But the next year the triplets cumulative age would be 48, and the mother would only be 47. So the situation would never occur.

Solution: There is a 50:50 chance that this will be possible.

(But see below for a more likely solution).

2. Rob Ferguson 1 June 2020 at 1:33 pm

My reading of the puzzle is that, although ages are measured as whole years, the date that the condition is met can be any day date in the future, not just the triplets’ birthday. In other words, we are seeking a solution for:

$3 \lfloor k \rfloor = n + \lfloor k \rfloor$

Therefore if the mother gave birth at 31, assuming that she did not give birth on her own birthday, on the day her triplets turn 16 years old, she will be 47, but will turn 48 before they turn 17. The triplets’ cumulative ages will then equal the mothers, satisfying the equation.

The possibility that the triplets’ cumulative ages equal the mothers will therefore be close to 100%, the only case where the equation isn’t met is if she gave birth to the triplets on her own odd-aged birthday.

• Jim Randell 1 June 2020 at 2:31 pm

I think you are right that I was only considering the case when their ages change at the same time (so, if the triplets are born on the mothers birthday, or we only do the comparison on the triplets birthday).

In general the ages would be interleaved, so we would get a sequence of birthdays like this:

M=30, T=0, M=31, T=3, …, M=44, T=42, M=45, T=45, M=46, T=48, M=47, T=51, …
M=31, T=0, M=32, T=3, …, M=45, T=42, M=46, T=45, M=47, T=48, M=48, T=51, …
M=32, T=0, M=33, T=3, …, M=46, T=42, M=47, T=45, M=48, T=48, M=49, T=51, …

And I was only picking up the M/T cases, and missing the T/M cases.

So, yes, in general there is always a time when adjacent values are the same age, except (as you point out) in the case where the birthdays are all the same. Then there is only a 50:50 chance.

So the required answer is probably “almost 100%”. (I make it about 99.86%, if birthdays are uniformly distributed).

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