# left and right hand sides are different why?

Let's say I have 50 coins total.

I give 20 coins to my friend

I am left with 30

Then I give 15 again

I am left with 15

Again I give 9

I am left with 6

Again I give 6

I am left with 0

When I sum them up back what I had left each time: 0+6+15+30 = 51

But I had only 50 rupees.

And when I calculated what I gave each time it is 6+9+15+20 = 50

And they both are different. Why they are different?

Because you should

actually add what you give out, not you have left each time

which is

20+15+9+6=50

Second question:

imagine this. you have 10 dollars. you spend them one by one. by adding what you have left each time you get 9+8+7+6+5+4+3+2+1=45 dollars. but you actually have 10 dollars in the beginning, not 45.

• What is this ??? – Hamza.S Feb 3 '19 at 16:47
• press on the yellow boxes. you will see things. these are spoilers, to warn people as some people may not want to read these @Hamza.S happy puzzling ;) – Omega Krypton Feb 3 '19 at 16:53
• Ok . Happy puzzling but it doesnot solve my answer – Hamza.S Feb 3 '19 at 16:55
• @Hamza.S the tags are fine. edited answer. hope this helps. if it does you can upvote by pressing the upwards arrow or accept the answer by pressing the check next to my answer ;) – Omega Krypton Feb 3 '19 at 16:59
• You can also argue, that you should add the 50 coins from the begging, giving you 101 coins. – yunzen Feb 3 '19 at 19:17

I'm not sure if this riddle is "closed" or something, but I'll try to back up @OmegaKrypton's one, since I came up with the same ending conclusion.

Basically:

Adding up what you're left with each time just doesn't "mathematically" work. In fact, while the sum of what you gave each time inevitably adds up to your initial total, the sum of what you are left with works differently.

Let's make some examples:

Since this is valid for every N total, we can use a smaller number to clarify better: let's pick 5. If we apply the same rules:
Give 2, left with 3. Give 1, left with 2. Give 1, left with 1. Give 1, left with 0. Summing up, the "given total" adds up to 5, of course. But the "left total" sums up to 6.

Now, let's try this setup.

Give 2, left with 3. Give 1, left with 2. Give 2, left with 0. Now the sums, are both equal to 5. This means that, while the total of what you give is always "correct", the sum of the values you are left with, depends on how you distribute the starting value over time. In other words, according on how many steps you take to make 50 reach 0 (and on how much you take away from 50), the value you'll get will inevitably change.

Hope this is worth to read!

First you start with 50 rupees, and little by little you give your money away without taking anything until you have nothing left. That means you give 50 in total (all the money you started with), but it has nothing to do with the sum of the amounts left. You can give away 1 rupee in every step until the last and get an even higher result. Letting the given amounts be $$g_1,g_2,...,g_n$$, the remaining amount in a particular step is basically $$50-(g_1+g_2+...+g_x)$$, and their sum doesn't necessarily equal 50 (in other words $$g_1+g_2+...+g_n$$).