Answer to Question #107193 in Algebra for Akshat Patel

Question

Puzzle:
Mayur has decided to offer five coconuts to Tirupati. To reach to Idol, he has to pass through 10 gates. At each gate, he has to offer one coconut for each bag he carries. Each bag can hold only 10 coconuts. How many coconuts should he carry from the main gate?

Answer 1

Let total number beg=x .

Then total number of coconuts$=10x$

Which is a even number.

To reach the idol ,he has to pass through 10 gates and at each gates he has to offer one coconuts for each bag he carries. And finally ,when he has reached in front of Ideal ,he have 5 coconuts in his beg.

which is a odd number .Since, when he loss 10 coconuts ,he also loss one beg because a beg have 10 coconuts.

Observe that ,when Mayur pass from one gates to other gates ,the number of coconuts he offers equal to number of begs he had at that time .

Now, initially he has $10x$ coconuts and finally he has 5 coconuts.

Therefore, in the last 5 gates he give one one coconuts .

i,e, at 6th gate he has only one beg.

Because 5 is an odd number and sum of all coconuts including 5 is an even number .

Therefore, the number of coconuts in the first 5 gates offered by Mayur $=10x-10(=$ sum of coconuts in the last gates+ 5$)$

$=$ multiple of 10 with only one beg left at 6th gate.

Therefore at 5th gate Mayur had 2 begs.

Hence ,$x=2$ and total number of coconuts = 20.

Note: Think from backward.

Diagrammatically,

$2\rightarrow2\rightarrow2\rightarrow2\rightarrow2\rightarrow1\rightarrow1\rightarrow1\rightarrow1\rightarrow1\rightarrow5$

The last palace is the in front of Idol.