In $1952$, I was as old as the number formed by last two digits of my birth year. When I mentioned this to my grandfather he surprised me by saying that the same applied to him also. The difference in our ages is ?

Now the way it has been solved is that let us assume that I was born in the year $19xy$ and my grandfather was born in the year $18pq$. Now as per the question :-
$19xy+xy=1952$ where $xy$ is my current age.
$\Rightarrow 1900 + xy + xy = 1952$
$\Rightarrow xy = 26$

$18pq+pq=1952$ where $pq$ is my grandfather's current age.
$\Rightarrow 1800 + pq + pq = 1952$
$\Rightarrow pq = 76$

Difference between the ages = $76 - 26 = 50.$

Now you all can see that this solution applied little logic with the assumption of my age and my grandfather's age. But what if the year given would have been $1989$ or $1999$ then we can't just assume that I would have been born in some $19xy$ and my grandfather would have been born in some $18xy$. It can very well happen that both of us would have been born in the same century.
I tried solving this by assuming that I was born in the year $abcd$ and my grandfather was born in the year $pqrs$ and following the same approach but it didn't helped me and then I turned to this solution that I have put up above.
So can anyone help me with this? Is it not possible to solve this problem by assuming years like I have or is it like we have to use some logic while assuming the birth years as the solution has done?

Thanks in advance !!!

  • 4
    $\begingroup$ "It can very well happen that both of us would have been born in the same century" — No, because there is a unique solution in the same century, and both cannot have the same birth year. Also, since the ages are both $\lt 100$ the only possibility remains that the subject was born in the current century, while the grandfather was born in the previous century. In that case a solution exists iff the current year is even, and in that case the difference is always $50\,$. $\endgroup$
    – dxiv
    Jul 3, 2021 at 6:48
  • $\begingroup$ @dxiv : have to say : +1 (also) - very elegant. $\endgroup$ Jul 3, 2021 at 7:32

2 Answers 2


Expanding on @dxiv comment, the existence of a solution depends on the parity of the reference year, but if there is a solution, then it is necessarily $100/2 = 50$. Let us see this in more details.

Let $r$ be the reference year ($1952$ in your original question). Following your notation, let $x$ and $y$ be the last two digits of your birth year, and $p$ and $q$ the last two digits of your grandfather's birth year. As you already observed, you get the equations $$ 1900 + 2xy = r \text{ and } 1800 + 2pq = r. \qquad (*) $$ Subtracting the second one from the first, one obtains $(1900 - 1800) + 2(xy - pq) = 0$, whence $pq - xy = 100/2 = 50$, independently of the choice of $r$.

Note that the previous computation does not require to know your age. However, to prove the existence of a solution, you need to verify that the equations $(*)$ have a solution and this requires the parity of $r$. For instance, if $r = 1999$, the equation $1900 + 2xy = 1999$ has no integer solution.

Now, to answer your question, you indeed need assumptions on your age and your grandfather's age. If you assume that you are born during the same century as your grandfather, the equations $(*)$ should be changed to $1900 + 2xy = r$ and $1900 + 2pq = r$, leading immediately to $xy = pq$, which is not possible.

Finally, I invite you to think of similar problems such like this one:

In year $r$, I was as old as three time the number formed by the last two digits of my birth year. Then my grandfather said that the same applied to him also. What is the difference in our ages?

In this case, the equations would be $1900 + 4xy = r$ and $1800 + 4pq = r$, leading to $(1900 - 1800) + 4(xy - pq) = 0$ and finally $pq -xy = 100/4 = 25$, a solution independent of the choice of $r$, but you would need $r-1900$ to be divisible by $4$ (like $1980$) to have a solution.

  • $\begingroup$ Thanks for such a good explanation. This cleared a lot of my doubts. Very well explained !!! $\endgroup$
    – Ganit
    Jul 3, 2021 at 16:30
  • $\begingroup$ You're welcome. $\endgroup$
    – J.-E. Pin
    Jul 3, 2021 at 16:57

I don't think, such an interesting scenario will occur if the mentioned year is $1999$ or $1899$. To show this, I will tell you a different approach to solve the problem-

Since in $1952$, I was as old as the number formed by last two digits of my birth year, I have to be $\frac{52}{2}=26$ years old in $1952$. If the same holds for my grandfather, he has to be not $26$, but $\left(\frac{100}{2}+26\right)$ (shifting by a century) or $76$ years old. Now, you see why 1999 or 1989 can't satisfy the property- it's because they're odd and you can't divide them by $2$.

Does that help?

Note that I assumed that a man cannot be more than $100$ years old (which is a valid assumption I guess :)). If you assumed infinite age, you can go on shifting centuries and getting more and more solutions.

  • $\begingroup$ Can anybody please explain the reason for this downvote? $\endgroup$ Jul 3, 2021 at 9:02
  • $\begingroup$ Sorry to say, but your post is simply wrong (there is no solution for the year 1999, since 1999 is odd). Furthermore, you don't really answer the OP question. $\endgroup$
    – J.-E. Pin
    Jul 4, 2021 at 8:55
  • $\begingroup$ @J.-E.Pin Can you please elaborate... I think the OP question was whether we will have a solution for the years $1999$ or $1899$. I clearly said, it won't, the proof of which in the explanation I provided. You can't divide $1999$ or $1899$ by $2$ because they are odd. Isn't that enough? $\endgroup$ Jul 4, 2021 at 9:08
  • $\begingroup$ I will not discuss with you any further, but unfortunately none of the words "even" or "odd" appear in your answer. $\endgroup$
    – J.-E. Pin
    Jul 4, 2021 at 10:21
  • $\begingroup$ @J.-E. Pin I thought the even odd part was trivial since a division by $2$ was involved. And I don't understand why you don't want to discuss any further. But anyways, thanks for suggesting the edit. $\endgroup$ Jul 4, 2021 at 11:20

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