when: bedtime

J1 wanted to return to our discussion about their three ages. In particular, someone had noticed that, prior to the recent birthday, their whole year ages were all primes: 3, 5, and 7. J1's question this evening: will this ever happen again?

# First case

We talked through 2 different cases. The first uses 2 year gaps, so AgeY(J3)+2 = AgeY(J2) and AgeY(J2)+ 2 = AgeY(J1). As we worked through examples, J1 quickly saw that it wouldn't work whenever J3's age was even, so we focused on odds. From 5, 7, 9, he realized that we could skip the cases where J3 was 7 or 9, so we looked at 11 and 17. Along the way, we had identified that J1's age was 9 (3x3), then 15 (3x5), then 21 (3x7). Seemed to be a pattern of always being a multiple of 3...# Second case

Next, we looked at their current gaps, so AgeY(J3)+2 = AgeY(J2) and AgeY(J2)+ 3 = AgeY(J1)This one was easier, solved by just looking at the parity of the ages.

So, sad news all around: they will never again have all prime ages!

# Fractional ages

Next, we talked about the equation Age(J3) + Age(J2) = Age(J1). This works out now when we use whole year ages (or using the floor function, if you prefer). There had been some chatter earlier about fractional ages, so J1 wanted to check that. These were the estimates J2 had given earlier for their ages:- J3: 3 1/4
- J2: 5 1/2
- J1: 8

We talked for a while and J1 realized that, as time passes, the sum of the sibling's ages increases twice as fast as his age does. That gave him a clue that it won't happen in the future, but did in the past. So, when was that?

Keeping in mind the pattern he'd recognized about changes, he guessed it was 3/4 of a year ago. Checking through was a good exercise in fractions and confirmed his conjecture.

From that, he wanted to get more precise about when that date was, 3/4 of a year ago, and then started talking about that having been a really special time for the three of them (unrecognized at the time, of course). When he started getting that precise, though, I offered that J2's estimates were a bit off and substituted J3 = 3 1/6, J2 = 5 5/6, and J1 = 8. This gave an even more satisfying conclusion once he worked out this case.

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