# Solution

Let's say I start off with "x" number of flowers from the garden. I dip them all in P1 and get "2x" number of flowers.

I offer "y" flowers in the first temple T1. So, I have 2x-y flowers remaining with me.

I dip these 2x-y flowers in the second pond P1. I get 2(2x-y) = (4x-2y) flowers. I offer "y" flowers in the second temple T2. So,after exiting T2 I have with me [(4x-2y) - y] = (4x-3y) flowers.

I dip these flowers in pond P3 and I get 2(4x-3y) = (8x-6y) flowers. I offer all these flowers in temple T3, and they are equal to "y" (because I have offered all the flowers with me, and I must offer the same number of flowers, "y", as in T1 and T2).

Therefore, (8x-6y) = y.

So, 8x = 7y.

Thus, x = 7y/8.

I can have only whole number of flowers (I can't have, for instance, 2.66 flowers!). I can have whole number of flowers only when "y" is 8 or a multiple of 8.

When I put y = 8, I get x = 7. Thus (7, 8) is a solution. Let's work this out: I pluck 7 flowers from the garden. I dip them in P1 and obtain 14 flowers. I offer 8 flowers in T1 which leaves me with 6 flowers. I dip these 6 flowers in P2 and I get 12 flowers. I offer 8 flowers in T2 and, so, I have 4 flowers remaining with me. I dip them in P3 and get 8 flowers and I offer them all in T3.

There can be an infinite number of solutions: if y = 16, x = 14; if y = 24, x = 21; if y = 32, x = 28; and so on. Therefore the possible (x,y) pairs are (7,8), (14,16), (21, 24), (28, 32), ..........

• Union Public Service Commission - www.upsc.gov.in
• IIT-Kharagpur - www.iitkgp.ac.in
• Indian Statistical Institute - www.isical.ac.in
• Indian Institute of Technology Madras - www.iitm.ac.in
• Indian Institute of Management, Ahmedabad - www.iimahd.ernet.in
• Indian Institute of Mass Commission - www.iimc.nic.in
• IIT Bombay - www.iitb.ac.in