Can you solve it? – Only One IAS in out of 3lac guys solved it
( ) + ( ) + ( ) + ( ) + ( ) = 30
This is what you have for equation. The following are the numbers that you can use to fill in the brackets:
1, 3, 5, 7, 9, 11, 13 and 15
You can repeat the numbers if required. The resulting sum should be 30.
This question appeared in the UPSC final examination in December 2013. Out of all the people who appeared, only one guy was able to solve this namely Gaurav Agarwal.
Can you solve it?
Remember we told you that Birbal is a witty trader. So his sole motive is to get rid of the sacks as fast as he can.
For the first sack:
He must be able to fill fruits from one sack to other two sacks. Assume that he is able to do that after M check points. Now to find M,
(Space in first sack) M + (Space in second sack) M = (Remaining fruits in Third Sack) 30 – M
M = 10
Thus after 10 checkpoints, Birbal will be left with only 2 sacks containing 30 fruits each.
Now he must get rid of the second sack.
For that, he must fill the fruits from second sack to the first sack. Assume that he manages to do that after N checkpoints.
(Space in First Sack) N = (Remaining fruits in second sack) 30 – N
N = 15
Thus after he has crossed 25 checkpoints, he will be left be one sack with 30 fruits in it. He has to pass five more checkpoints where he will have to give five fruits and he will be left with twenty five fruits once he has crossed all thirty check points.
The main issue here is that an addition of odd numbers for odd times will result in an odd number. So in no way will you get an even number like 30 by adding these odd numbers.
So you will have to think out of the box to solve this one.
Use of mathematical operator is a must.
To get an even number from an odd number the best operation to use would be factorial.
So we use 3!,
3! = 3 x 2 x 1 = 6
so if we add 3! five times we get the answer as 30. (As repetition of the number is allowed.)
Hence the answer would be;
3! + 3! + 3! + 3! + 3! = 6 + 6 + 6 + 6 + 6 = 30.
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