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A group of people have gathered for a birthday celebration. Their ages are related as follows:

  • The product of the 1st person's and the 2nd person's ages is $311\frac{2}{3}$ plus the 3rd person's age.
  • The difference between the 1st person's and the 2nd person's ages is $2\frac{31}{33}$ times the 3rd person's age.
  • The quotient of the product of all their ages and the sum of all their ages is $826\frac{4}{29}$.
  • The sum of the 1st person's age and the quotient of the 3rd person's and the 2nd person's ages is $41\frac{17}{24}$.
  • The square of the 3rd person's age is triple the 1st person's age.

Whose birthday is it? And what is each person’s age?

Hint:

Consider the number of relationships given.

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  • 3
    $\begingroup$ @bobble From the link, "A good puzzle can be pure maths, i.e. it is perfectly possible, that e good puzzle requires pure maths to be solved. The puzzle can be that one has to find the mathematic which is needed. Or it might be, that the maths involved is unexpected." I think this fits that description. $\endgroup$ – asg Jul 29 at 22:12
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    $\begingroup$ I think the question "Whose birthday is it?" should have been emphasised (leaving figuring out of the ages as an intermediate step). Figuring out whose birthday it is is definitely not "pure maths" and makes this more of a puzzle. Sadly, I have no power to suggest reopening. $\endgroup$ – Earlien Jul 30 at 2:18
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    $\begingroup$ I think this should be reopened it looks to me a bit like priest and vicar question in which we have to figure out age of priest. It's more of a puzzle style and requires ingenuity to solve rather than maths problem. Sadly I can't reopen this. $\endgroup$ – Lakshay Sura Jul 30 at 7:14
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    $\begingroup$ @asg One is enough. If four people downvote you and all four leave a comment, it will feel like we're piling on, or ganging up on you. That's one of the (many, many) reasons why people are not required to explain their downvotes and in some cases are actually discouraged from doing so. I've voted to re-open the question; thank you for taking our feedback on board. $\endgroup$ – F1Krazy Jul 30 at 8:01
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    $\begingroup$ I'll vote to reopen because I believe it can be answered without any detailed calculations. P.S. "more than" in maths is usually ambiguous because it can imply addition or multiplication. Please can you be explicit. $\endgroup$ – chasly - supports Monica Jul 30 at 11:39
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Let us denote the ages of Person 1, Person 2, Person 3 by $x,y,z$ respectively. We'll assume that $x,y,z$ are positive throughout.

The product of the 1st person's and the 2nd person's ages is $311 \frac{2}{3}$ plus the 3rd person's age.

$$xy - z = 311 \frac{2}{3} = \frac{935}{3}$$

The sum of the 1st person's age and the quotient of the 3rd person's and the 2nd person's ages is $41 \frac{17}{24}$

$$x + \frac{z}{y} = 41 \frac{17}{24} = \frac{1001}{24}$$ $$\Rightarrow xy + z = \frac{1001}{24}y$$

Subtracting the first equation from the second gives

$$2z = \frac{1001}{24} y - \frac{935}{3} \Rightarrow z = \frac{1001}{48}y - \frac{935}{6}$$ $$\Rightarrow xy + \frac{1001}{48} y - \frac{935}{6} = \frac{1001}{24} y$$ $$\Rightarrow x = \frac{1001}{48} + \frac{935}{6y}$$

The difference between the 1st person's and the 2nd person's ages is $2 \frac{31}{33}$ times the 3rd person's age.

$$x-y = 2 \frac{31}{33} z = \frac{97}{33} z = \frac{97097}{1584}y - \frac{90695}{198} = \frac{8827}{144} y - \frac{8245}{18}$$ $$\Rightarrow x = \frac{8971}{144}y - \frac{8245}{18} = \frac{1001}{48} + \frac{935}{6y}$$ Multiplying across by $144y$ and rearranging gives $$\Rightarrow 8971 y^2 - 68963 y - 22440 = 0$$ Solving the quadratic equation for $y$ gives $$ y = \frac{68963 \pm \sqrt{4755895369 + 805236960}}{17942} = \frac{68963 \pm \sqrt{5561132329}}{17942} = \frac{68963 \pm 74573}{17942}$$ Note here that calculating the square root is the trickiest part of this whole calculation but if we notice that our given number is a little less than $56 \times 10^8$ then we should expect the result to be a little less than $7.5 \times 10^4$. Using some finer estimation like this allows us to hone in on the answer a little quicker and we can use a binary search to narrow down to the given value.
Using the assumption that $y$ be positive, we find that $$y = \frac{143536}{17942} = 8$$ and substituting into the equations for $x$ and $z$ we find that $$ x = \frac{1001}{48} + \frac{935}{48} = \frac{1936}{48} = 40 \frac{1}{3}\,\,\,\,,\,\,\,\, z = \frac{1001}{6} - \frac{935}{6} = \frac{66}{6} = 11$$

Now let us check the consistency with the other equations

The square of the 3rd person's age is triple the 1st person's age

$$3x = 3\left(\frac{121}{3}\right) = 121 = 11^2 = z^2$$ so this is consistent

The quotient of the product of all their ages and the sum of all their ages is $826 \frac{4}{29}$

A quick check is enough to convince us that $$ \frac{xyz}{x+y+z} \neq 826 \frac{4}{29}$$ However, the question does not specifically state that there are only $3$ people at the party so let us assume there is a fourth person, Person 4 whose age is $w$. Then we require $$ \frac{xyzw}{x+y+z+w} = \frac{10648w}{178 + 3w} = \frac{23958}{29}$$ $$\Rightarrow 308792w = 4264524 + 71874w$$ $$\Rightarrow w = \frac{4264524}{236918} = 18$$

Whose birthday is it?

Since theirs are exact ages it must be the birthdays of Person 2, Person 3 and Person 4 and they are $8, 11$ and $18$, respectively

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    $\begingroup$ nicely done -- but I wouldn't call this "brute force" because brute force usually implies guess and try over a large range of possibilities. This is (tedious) algebra plus a key insight (rot13: gur sbhegu crefba). $\endgroup$ – Ross Presser Jul 31 at 15:23
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    $\begingroup$ In addition to solving the puzzle, I’m also impressed by your skills in typing out equations. I gotta learn that stuff as well as the puzzle solving skills. $\endgroup$ – asg Jul 31 at 15:27
  • $\begingroup$ @RossPresser Yeah, you're right. I've taken that line out now. $\endgroup$ – hexomino Jul 31 at 16:08
  • $\begingroup$ rot13( V guvax gurer pbhyq or zber guna sbhe crbcyr, ohg vg vf nzovthbhf. ) $\endgroup$ – tehtmi Aug 5 at 4:06
  • $\begingroup$ There's no need to rot13(hfr n ovanel frnepu gb trg gur fdhner ebbg); there are ways to calculate this directly with pencil and paper. $\endgroup$ – shoover Aug 9 at 17:34

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