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.

  • $\begingroup$ Fun fact: if two people live long enough there will always be exactly a year during which one of them is twice as old as the other... $\endgroup$
    – Dr Xorile
    Commented Mar 7, 2019 at 7:33
  • $\begingroup$ Apart from First three lines, any other lines has anything to do with answers? I think we can get the answer using only first three lines. $\endgroup$
    – gopal
    Commented Mar 7, 2019 at 10:06

2 Answers 2


I think that Bob is



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.

  • $\begingroup$ Looks like there are different solutions to the problem $\endgroup$
    – Jerry
    Commented Mar 7, 2019 at 10:34
  • $\begingroup$ This is indeed the minimal. Playing around with the order of birthdays you can make Bob's age have quite a range. Judging by your proof I had a different method of solving the problem in mind than what you did. Perhaps next time I'll change dad to granddad and ask for the second highest age Bob can be.. Lol. Thanks for trying the problem out for me! $\endgroup$
    – Amorydai
    Commented Mar 7, 2019 at 12:49
  • $\begingroup$ @Jerry he did say that bob is the youngest possible with the given information, so should only be one solution $\endgroup$
    – Quinn
    Commented Mar 7, 2019 at 14:16
  • $\begingroup$ @Quinn Yes, now I know. OP commented on my answer earlier, it was my bad $\endgroup$
    – Jerry
    Commented Mar 7, 2019 at 14:24

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.

  • $\begingroup$ I also tried similar approach.. HaHa $\endgroup$
    – gopal
    Commented Mar 7, 2019 at 10:08
  • 1
    $\begingroup$ I think it makes more sense to say Bob is 3 when they moved in and "not yet 7" when the house burned. Although I agree with your approach, it cannot be the case that I am, at one time, 4 years older than Bob and then 6 years older at another time. The variation is at most 1. $\endgroup$
    – hexomino
    Commented Mar 7, 2019 at 10:11
  • $\begingroup$ @hexomino Oops, that's a valid point! $\endgroup$
    – Jerry
    Commented Mar 7, 2019 at 10:17
  • $\begingroup$ @hexomino Actually, I had something else wrong lol. I had 18y (Bob) - 16 years ago = 3 when it should be 2, so the months will actually be reversed. $\endgroup$
    – Jerry
    Commented Mar 7, 2019 at 10:25
  • 1
    $\begingroup$ @Jerry In the problem I stated Bob was the youngest he could be given the statements. Thank you for trying my problem out! $\endgroup$
    – Amorydai
    Commented Mar 7, 2019 at 12:43

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