Dtjytyk

I created this problem myself, but I did have inspiration.

Today is a special day, I'm moving out of my parent's house. My father comes to me and says "Wow, you're half my age. On the day we moved into this house, I was four times your age, and your brother Bob was 3 years old." To which I reply "Good times. I remember the day when I almost burned the house down, you were three times my age, and I was twice as old as Bob."

My father then asks "How old is Bob anyway?" We think about it a minute and then my father's eyes bulge out "Bob can't be 27! I am not losing my mind, am I?"

"No", I reply, "Of course Bob is younger than 27. Actually, he is the youngest he can be given what we've already said". How old is Bob?

A useless hint:

As by convention, a person is said to be X years old, if he has lived at least X years, but has not yet lived X+1 years. A person's age is an integer. If I say someone is three times my age, and my age is X, then their age is 3X.

I think that Bob is

$15$

Reasoning

Suppose my birthday is on the $6$th of March and today is the $7$th.
If we moved in on the $5$th of March $12$ years ago, Bob is $3$, I am $7$ (not yet turned $8$), Dad is $28$.
When the fire happened (say $7$th of March $10$ years ago) Bob was $5$, I was $10$ and Dad was $30$.
Today, Bob is $15$, I am $20$ and Dad is $40$.

Proof that this is minimal

I must be older than Bob. If I am younger than $7$ when we moved house then I must have been either $4$, $5$ or $6$. If I was $4$ then I have already passed the point of being double Bob's age. If I was $5$, then I will be double Bob's age only if my birthday is about to occur before Bob's ($5$ to $6$). In that case, Dad goes from being $20$ to being $18$ so that doesn't work.

Therefore, the only other possibility is that I was $6$ when we moved house. In that case, Dad was $24$ and must always be either $17$, $18$ or $19$ years older than me. To be triple my age when the house burned down, the only possibility is that I am $9$ and Dad is $27$ but this leaves no option for Bob since $9$ is odd.

It was more trial and error:

If today Bob is 18, I am 24 and father is 48 (current age of father being twice mine is held)

And if 12 years ago the house burn occurred, that made Bob 6, me 12 and father 36 (ratio Bob:me:father is 1:2:3)

And if 16 years ago the move occurred, that made Bob 2, me 8 and father 32 (age of father being 4 times mine)

Then it works out, so Bob is 18.


Now people are not all born on the same date, so sometimes there is a small difference in age. Let's say Bob's birthday is in June, mine in August and father in November. If we are in May and the house burn is also in May, then the birthdays don't matter too much. However, if the move happened in July, it means Bob was 3 already at the time of the move.


The year the house burn occurred, the ratio is a clear 1:2:3, which meant that it must have been a 6n year from now, and for the move, the ratio of ages we had is 1:4, we it must have been a 4m year from now, where 4m > 6n. I tried n=1, m=2, so Bob would be at least 17, but this put father at 66y the year of the house burn, 72y today, and me 36y today, which doesn't add up by more than a year (I would consider 1 year difference fine for the reason I specified in the previous paragraph). I went on trying until n=2, m=4.