Let's say that I have 50 coins, total.

  • I give 20 coins to my friend; I'm left with 30.

  • Then I give 15 more; I'm left with 15.

  • I give 9 more; I'm left with 6.

  • I give 6; I'm left with 0.

When I sum-up what I have left each time, I get $$0+6+15+30 = 51$$ But I only had 50 coins.

When I sum-up what I give every time, I get $$6+9+15+20 = 50$$

The sums both are different? Why are they different?

  • 1
    $\begingroup$ why should these be the same? $\endgroup$ – Lord Shark the Unknown Feb 3 at 17:08
  • $\begingroup$ why they are different ?? $\endgroup$ – Hamza.S Feb 3 at 17:08
  • 2
    $\begingroup$ See en.wikipedia.org/wiki/Missing_dollar_riddle . Neither that, nor this, are serious maths problems. $\endgroup$ – Lord Shark the Unknown Feb 3 at 17:15
  • 3
    $\begingroup$ If you gave your first friend $50$ then you'd have left with $0$, so that sum would just have been $0$...which is also not the same as $50$. $\endgroup$ – lulu Feb 3 at 17:48
  • $\begingroup$ @Lord Shark the Unknown: otherwise ask an accountant (20+15+9+6=50). $\endgroup$ – cgiovanardi Feb 3 at 23:01

There's absolutely no reason for the sums of the "leftover amounts" to be equal to what you started. For a simple example of this, consider what happens if each day you give away one coin. Then your "leftover amounts" are $49,48,47,46,...$ and their sum is a bit more than $50$ (precisely, it's $1225$). Conversely, if you give away everything at once, you only have one leftover amount and it's a bit less than $50$ (precisely, it's $0$).

The point is that in order to expect the sum of the "leftover amounts" and the original number of coins to be the same, you'd need to expect that each coin is left over at exactly one moment in time (so each coin gets represented exactly once in the sum of the "leftover amounts"). But that's obviously nonsense in general.

On the other hand, each coin you start with is eventually taken away, and you can never take away the same coin twice - so that's why the sum of the "takeaway amounts" gives the number of coins at the start (each coin is counted exactly once in exactly one of the "takeaway amounts").

Basically, before you expect to add up a bunch of numbers and get a particular result, you need to have some reason to believe that that collection of numbers and that expected result are related. And you don't have one here.


Your first calculation summed what remained at various stages. There's nothing special about that sum in general.

Say we start with $N$ rupees, so in your example $N=50$. Now subtract $a_1$ of them, $a_2$ of them etc., until the last $a_k$ of them are subtracted. In your example $$a_1=20,\,a_2=15,\,a_3=9,\,a_4=6,\,k=4.$$So, these numbers sum to $50$. But the numbers of rupees remaining after each giveaway are $N-a_1,\,N-a_1-a_2,\,\cdots,\,0$. These sum to $$\sum_{j=1}^k\left(N-\sum_{l=1}^j a_l\right)=kN-\sum_{l\le j\le k}a_l=kN-\sum_{l\le k}(k-l+1)a_l.$$Or to make the algebra less scarier for your example, we get $$4\times 50-4a_1-3a_2-2a_3-a_4=200-80-45-18-6=51.$$There's no reason you'd expect this to give $N$. Of course, writing $N$ as the sum of each subtracted number gives an alternative expression for what was $51$ in this case: $$\sum_l (l-1)a_l=a_2+2a_3+3a_4=15+2\times 9+3\times 6=51.$$

  • $\begingroup$ why not to add that ? What we left is not equal to what we have first ? $\endgroup$ – Hamza.S Feb 3 at 17:10
  • $\begingroup$ @Hamza_S The only way I can think of to explain what happens instead is with some algebra, as in my edited answer. I hope it's helpful. $\endgroup$ – J.G. Feb 3 at 17:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.