Chocolate Theft

Algebra Level 3

3 boys stole a basket of chocolates and hid the chocolates.They counted the chocolates which were less than 100. Then they went to sleep.In the night the first boy woke up and saw that if he ate 1 chocolate then the remaining could get divided into 3 equal parts. So he ate 1 and took 1/3rd of the remaining and kept it separately. Then the second boy woke up. He also ate 1 chocolate and took 1/3rd of the remaining and kept it separately.Then the third boy woke up and followed the same procedure.In the morning when they counted then they observed that the remaining chocolates were too in the form $3n+1$ . Find the original number of chocolates stolen.

The answer is 79.

2 solutions

Ashwin K
Feb 10, 2016

Let number of initial chocolates = 3n + 1.

When first person eats his share, remaining will be = 2((3n+1) -1)/3 = 2n.

When second person eats his share, remaining will be = 2(2n-1)/3 = 4n-2/3.

When third person eats his share, remaining will be = 2(4n-5/3)/3 = 8n-10/9.

By using remainder concept, 8 after multiplied with n should carry a remainder 1 (i.e)    8 carries a negative remainder -1 and so n also carries negative remainder -1.
Those numbers of that form are 8,17,26.
Final number also of form 3n + 1 which gives us unique answer with n =26.

Hence, the answer is $\boxed{79}$ .

Lakshya Singh
Feb 10, 2016

Ans 79

Let boys be A,B,C

A- 79-1=78/3=26

B-52-1=51/3=17

C-34-1=33/3=11

Mangoes in the morning=33-11=22

Therefore 22 is in the form of (3n+1)

Chocolate Theft

The answer is 79.

2 solutions

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