Magical Prayers-- II

x is original number of flowers. Dipping in magical water triples to 3x. TemPLE 1: 3x-y is flowers leftover after offerings. Dip in water : 3(3x-y) is what she has for temple 2. Temple 2: 9x-3y-y=9x-4y are flowers leftover after offerings. Dip in water: 3(9x-4y) is what she has for temple 3. Temple 3: 27x-12y-y are flowers leftover after offerings. Dip in water: 3(27x-13y) Finally, temple 4: 81x-39y-y=0 Nothing leftover. 81x-40y=0 81x=40y, x/y=40/81 She has 40 flowers in the beginning and she gave 81 flowers at each temple.

A good candidate for mental math

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What ever be the count of the flowers he offered in the last temple was because it was tripled 4 times in a row i.e. $3^4=81$ . So, 81 flowers were offered to the last temple as well as all the previous temples he visited. Now to determine the initial count, we have to back calculate. But think in the other way, what ever be the count, these flowers were shared in the ratio of 3:9:27:81 in the subsequent temples he visited, so we can safely say, that the total number of flowers he was originally caring was

$\frac{81}{3}+\frac{81}{9}+\frac{81}{27}+\frac{81}{81}=27+9+3+1=40$

So he was caring 40 flowers originally delivering 81 flowers in each visit

Ashwini starts with $a$ flowers and offers $b$ flowers at each temple.

Before arriving at Temple 1, the number of flowers that Ashwini has ( $a$ ) is tripled.

$3a$

After leaving Temple 1, $b$ is subtracted from $a$ .

$3a-b$

This process is repeated on the trip to Temple 2,

$3(3a-b)-b=9a-3b-b=9a-4b$

to Temple 3,

$3(9a-4b)-b=27a-12b-b=27a-13b$

and to Temple 4.

$3(27a-13b)-b=81a-39b-b=81a-40b$

After leaving Temple 4, Ashwini has 0 flowers left.

$81a-40b=0$

After rearranging:

$81a=40b$

$a=40$ and $b=81$ , so your final answer is 40.81.

Venture Hi has given the way to solve it. Just showing the pattern in this problem...

The pattern that emerges is flowers offered at each temple will be 3^ number of temples and flowers carried initially is minus 1 of that divided by 2..

2 temples -- 9 offered at each temple and 4 carried initially

3 temples-- 27 offered and 13 carried initially

4 temples -- 81 offered and 40 carried initially

general solution Let n temples and flower gets m times then min flowers (m^n-)/(m-1) and offering should be m^n.

If we map out the Formula for the equation we get. 3(3(3(3a-b)-b)-b)-b=0. Solve for a and we get a=b/81 + b/27 + b/9 + b/3, We know that the number of flowers given away at each site has to be divisible by 81, the Problem is Looking for the smallest number of flowers, 81 is the smallest whole numebr divisible by 81 so b=81. Then it is a simple matter of solving for a. (1+3+9+27)= 40=a 81=b

The last equation at the 4th temple = 81x-40y = 0 ,where x is the initial number of flowers and y is the number of flowers offered in the 4 temples.

So x= 40 & y= 81