# How many flowers to buy? [duplicate]

A man decided to place some flowers on three different temples. There is a magical pond before each temple and he has to swim through each one. So his trip will be:

[ pond1 - temple1 - pond2 - temple2 -pond3 - temple3 ]

The fact about magical ponds is that if anyone swims across the pond with flowers, the number of flowers will be doubled.

He buys some flowers and goes for his trip to the three temples.

At the end of trip, he has no flowers remaining and he has placed an equal number of flowers at all three temples.

How many flowers did he buy and how many did he place at each temple?

• This is a good old puzzle, and the generic version is gold :) Jun 13 '16 at 13:29
• Is this related to this question?
– f''
Jun 13 '16 at 13:42
• Doesn't he have to swim away from the temple, too? Jun 13 '16 at 14:31
• It doesn't matter where the ponds are, he just has to swim through them.
– Carl
Jun 13 '16 at 19:00
• 0 is a cheeky answer. Jun 14 '16 at 5:51

Let's say

x = number of flowers he brought
n = number of flowers he leaves in each temple

We get an equation:

$((x*2-n)*2-n)*2-n=0$

We get:

$8x -7n = 0$

Then

Smallest Common Multiple of $8$ and $7$ is $56$

So

We get $x=7$
He brought $7$ flowers

Step-by-step:

He brings $7$ flowers
He swims through $pond1$ and now has $14$
He leaves $8$ at $temple1$ and now has $6$
He swims through $pond2$ and now has $12$
He leaves $8$ at $temple2$ and now has $4$
He swims through $pond3$ and now has $8$
He leaves $8$ at $temple3$

• Alternatively it could be any multiple of 7 :) Jun 13 '16 at 13:29
• Quite - our man could buy any multiple of 7 flowers providing he leaves the same multiple of 8 flowers at each temple. Therefore it seems to me the puzzle is a bit weak, because it has multiple solutions (infinitely many, in fact, assuming our man has infinite money to buy the things, and infinite strength to tow them across the pond). Jun 13 '16 at 15:25
• A nice way to think about it is that he buys one flower and leaves it and all of its progeny at the third temple. Then to leave the same number at each he has to buy two for the second temple and four for the first, totaling seven. Jun 13 '16 at 18:43
• The smallest common multiple part is irrelevant to the answer. Jun 14 '16 at 16:10

He bought

7 flowers

and placed

8 flowers at every temple.

14 after pond1
6 after temple1
12 after pond2
4 after temple2
8 after pond3
0 after temple3

Another method to get the result:

Let's check it backwards. A man placed all the flowers he had.
Let's go even more back in time: Number of flowers halves three times so he placed a multiple of $2^3$ flowers in each temple (don't mind the addition here).
Now, let's find an initial number of flowers (when using the lowest possible number of flowers he placed in each temple):
$0 ← 8 ← 4 ← 12 ← 6 ← 14 ← 7$.

I know it's just the another method. But I think it's more handy. We can create a formula for a lowest number of flowers the person placed in each temple:

$$x=m^n$$
$m$ for multiplication of each magic pond and $n$ for a number of magic ponds.

Nice puzzle!
Here's another explanation of a solution:

We know that he has to leave an even number of flowers at the last temple, because he is leaving some number (n) which was just doubled.

Before pond 3, he has n/2.
Before temple 2, he has n/2 + n = (3/2)n.
Before pond 2, he has ((3/2)n)/2 = (3/4)n.
Before temple 1, he has (3/4)n + n = (7/4)n.
Before pond 1, he has ((7/4)n)/2 = (7/8)n.

The lowest n that [is divisible by all these denominators and therefore] makes all these values integers is 8, or more generally 2 raised to the number of ponds.

So he buys a minimum of 7.
After pond 1, he has: 14
After temple 1, he has: 6
After pond 2, he has: 12
After temple 2, he has: 4
After pond 3, he has: 8
After temple 3, he has: 0

Or, any multiple of 7, for example 14, leaving the same multiple of n.
After pond 1, he has: 28
After temple 1, he has: 12 (left 16)
After pond 2, he has: 24
After temple 2, he has: 8 (left 16)
After pond 3, he has: 16
After temple 3, he has: 0 (left 16)