Grandma Troll Problem

It is best to start with the last troll. If you need 2 cakes for grandma, and he gives you back 1, then 1/2 of the cakes = 1 cake, so you should get to the last troll with 2 cakes.

You can repeat this with the other 6 trolls to find that you only need 2 cakes to start with.

Here is how you can check your work mathematics:

With 2 cakes,

Troll 1: $2 ÷ 2 + 1 = 2$

Troll 2: $2 ÷ 2 + 1 = 2$

Troll 3: $2 ÷ 2 + 1 = 2$

Troll 4: $2 ÷ 2 + 1 = 2$

Troll 5: $2 ÷ 2 + 1 = 2$

Troll 6: $2 ÷ 2 + 1 = 2$

Troll 7: $2 ÷ 2 + 1 = 2$

And you have your $2$ cakes left for granny!

Mathematical way of doing the problem.

Let ${N}_{r}$ be the number of cakes I have after crossing $r$ bridges.

${N}_{0}=x\\{N}_{1}=\frac{x+2}{2}\\{N}_{2}=\frac{x+6}{4}$

Similarly,

${N}_{r}=\frac{x-2+{2}^{r+1}}{{2}^{r}}$ .

Therefore,

${N}_{7}=\frac{x+254}{128}$

Given ${N}_{7}=2$

$x+254=256\\x=2$ .

Therefore, I must carry 2 cakes.

I can not waste any cake,so I must take 2 cakes so that each time I give half of it(1) to the troll,no. of cakes remains the same, when he gives me one back.

2/2=1(I give half) 1+1=2(he gives one back) It remains the same.

Upvote please!!!!!!!!!

(Going backwards in time)

28) I give grandma $2$ cakes, after leaving Troll $7$ with them
27) Troll $7$ gives back to me $1$ cake
26) I had $1$ cake after Troll $7$ took half of what I had
25) Troll $7$ takes $1$ cake from me, half of the $2$ I had
24) I left Troll $6$ carrying $2$ cakes....
23)...(repeat)

Hence, there is no other solution other than to start the trip with $2$ cakes.

backtracking is the most optimal way to solve this.

at 7th bridge after giving away half of what u had...u r left with 2 cakes

at 6th bridge suppose u have x cakes x/2 +1 = no. of cakes at the 7th bridge...here
x=2...hence 2 cakes at the 6 th bridge too

at 5 th bridge , say u have x cakes again x/2 +1= no .of cakes at 6 th bridge x/2+1 =2 .x=2 ; ; . ; similarly 2 cakes at every bridge results in 2 cakes at the end

You need 2. But on your way home, you get 1 cake after the first bridge, then 1.5 after the second bridge, then 1 3/4 after the third, then 1 7/8 after the fourth, etc. If there were countably infinite bridges to cross, you'd get back home with 2 cakes. (As it is, you only have 1 and 255/256ths cakes.)

You only need two cakes; all the trolls are a bit thick! (or just too generous!)

Let's he started with A cakes. So, After the first Bridge, he left with {A-A/2 +1} = {(A + 2)/1} After Second Bridge = (A+6) /4 After third Bridge = (A+14/8) So, we can see a pattern after crossing each Bridge= {(A + 2^ (n+1) - 2) / 2} where n is no. of bridges. so after 7 th Bridge, he should have 2 cakes. solve the equation you will get the answer

ans is 2......

We need 2 cakes in the end, let the total no. of cakes be x, and we have 7 trolls which divide our no. of cakes by 2 and give one back, this implies the equation ((((((x/2 +1)/2 +1)/2 +1)/2 +1)/2+1)/2+1)/2+1= 2 Every time we subtract 1 from RHS we also multiply by 2, which gives us the final answer 2

This troll isn't greedy at all!

Since he'll give you back 1 of your cakes whenever you cross a bridge, and you give him a 1/2 of all your cakes when you cross a bridge, you should bring 2. It' basic logic.

1st Troll : 1 cake given, 1 cake received

You can repeat this with the next 6 trolls

Or, you could put poison in the cakes :)

Array yar ......every Indian solve this.

2 because when ever you give him half of your cakes he will give you a one back then your cakes are going to stay 2 for ever

Well every troll takes half of your cakes and gives u one more. So let's look at the last troll. How much cakes u need in order to have 2 cakes remaining? The answer is 2.... yeah have two and none of the trolls can have any net effect on ur number of cakes cuz they are so nice that if u havetwo cakes they give back what they took. So there it is- just go with two cakes and troll the trolls.

We know that in the end we will have two cakes. We go backwards. We give the troll back his one cake he gave us out of niceness(-1). The troll gives us back half our cakes he took from us (Now we have 2 times the amount as before. Hence times 2). Repeat 6 more times. This can be written in a programmable for loop in any programming language.

x=2

for (i=0; i<7; i++)

  - x=(x-1)*2

Another solution is a recursive function:

x=number of cakes we want to give grandma

y=number of bridges we have to cross

f (x, y) will be defined as followed:

If y=0 we have reached grandma so f (x,0) will be x

Otherwise we call the function again with following parameters: f ((x-1)*2, y-1)

So for the start we have f (2,7)=f ((2-1)*2, 7-1) And so on.

To solve this problem I thought backwards.

To have 2 cakes after crossing the last troll (#7!), how many cakes should I be carrying? The answer was 2: if I arrive with 2 cakes, I have to give him half of them (=1), and he gives me back 1 cake. This made me realize that, by going backwards, if I had two cakes with each troll I'd always keep all two of them.

working backwards, when you cross back to the 7th bridge, you subtract one cake:1, and then multiply that by two, so 2. it repeats so you need to start with 2 cakes.

Since it is half a cake, and the troll has to give you a full cake, you will get 2 cakes after crossing the bridge all 7 times. You start with half a cake , and then you get 1 cake back, so now you got 1 cake. Give half of that cake to troll on the next bridge, you would have 1.5 of a cake. Next bridge you would have 2. and so on and so on. The less cakes you take the more cakes you shall end up with. 2/2+1 is the equation.