So I came across this word problem and would like to get some help defining the model behind it and how to solve it.

There is a temple, whose premises have a garden and a pond. It has 4 idols, each of Ram, Shiv, Vishnu and Durga. The priest plucks x flowers from the garden and places them in the pond. The number of flowers doubles up, and he picks y flowers out of them and goes to offer it to Lord Ram. By the time he reaches to the pond, he finds the remaining flowers also have doubled up in the meantime, so he again picks up y from the pond and goes to Lord Shiv. This process is repeated till all the Gods have y flowers offered to them, such that in the end no flower is left in the pond. Find x and y.

I got as far as:

epoch0, x flowers
epoch1, 2(2x-y) flowers
epoch2, 2(2(2x-y)-y) flowers
epoch3, 2(2(2(2x-y)-y)-y) flowers
epoch4, 2(2(2(2(2x-y)-y)-y)-y) flowers

Then I thought of setting epoch4 to 0. But I know I need another equation since there are two unknowns. Also, I have a hunch I could also use an exponential growth equation but the constant "withdrawals" is something I have not seen before. Im kinda stuck and would appreciate any tips!

The answer is (base64 encoded): eD0xNSx5PTE2

  • $\begingroup$ There are infinitely many solutions, no? $\endgroup$ – uniquesolution Mar 14 '17 at 7:40
  • $\begingroup$ @uniquesolution yes there are multiple... but what would be the mechanics/method to even get one? Is my model correct? $\endgroup$ – nemo Mar 14 '17 at 18:03

You have a linear difference equation $x_{n+1}=2x_n-y$ where $x_0=x$

Let $z_n = x_n -c$ for some constant $c$ which we will determine.

Then $z_{n+1}+c=2(z_n+c)-y\implies z_{n+1} + c = 2z_n + 2c -y$

Setting $c=2c-y$ gives $c=y$ and, moreover, we get that $z_{n+1}=2z_n$

From this we get that $z_n=2^nz_0\implies x_n-y=2^n(x_0-y)$

Now we can plug in $n=4$ and use the fact that $x_4=0$ to solve for $y$ to get

$$-y = 16(x-y)\implies 15y=16x$$

Any whole numbers $x$ and $y$ that satisfy this equation will work.


It seems to me that if the priest plucks $15$ flowers from the garden and places them in the pond, upon which they double up, and then he picks $16$ out of them and gives them to Ram. When he comes back, there are $28$ in the pond. He picks $16$ again, and gives them to Shiv. He comes back, there are $24$. He picks $16$ again, and gives them to Vishnu, only to find $16$ flowers in the pond, which he gladly takes and hands over to the fourth lord, Durga. I came up with this solution by noticing that the series you wrote comes down to $16x-15y$ after the fourth step.


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