There are two gates. One goes to hell and the other goes to heaven. The gatekeeper asks a puzzle to Sahil in order to decide which gate he should open for him. Obviously if Sahil answers the puzzle correct , the gate to heaven will be opened else the gate to hell. According to the problem Sahil is given n consecutive integers from 1 to n ,which are written in a row. He has to put signs “+” and “-” in front of them so that the expression obtained is equal to 0 or, if the task is impossible to do,then Sahil should tell the gatekeeper “No solution exists for the given problem “.The gatekeeper expects Sahil to find his answer in minimum time using an efficient approach rather than examining all possible ways to place the signs. Sahil comes to you considering you as his friend. Would you help him out to get to the solution ?
The puzzle has a solution if and only if either n or n+1 is divisible by 4.
The problem is equivalent to partitioning n integers from 1 to n into two disjoint subsets with the same sum. Disjoint subsets are the subsets with no common element. The two disjoint subsets will be :
1) a subset of numbers with a plus sign before them , and
2) a subset of numbers with a minus sign before them.
Since the sum of first n consecutive integers is S = 1 + 2 + 3 + —————— + n = (n)(n + 1)/2 ,the sum of the numbers in each of the subsets must be equal to exactly one half of S, i.e, S/2. It means that the value of S, i.e, (n)(n+1)/2 must be even for the puzzle to have a solution.
1) It is clear that that n(n + 1)/2 is even if and only if either n is a multiple of 4 or n + 1 is a multiple of 4.
2) Indeed, if n(n + 1)/2 = 2t, this implies that n(n + 1) = 4t, and since either n or n + 1 is odd, the other must be divisible by 4. Conversely, if either n or n + 1 is a multiple of 4, and hence n(n + 1)/2 is obviously even.
3) Now considering the case when n is divisible by 4. In this case, we can, for example, partition the sequence of integers from 1 to n into n/4 groups of four consecutive integers and then put “+” sign before the first and fourth number and “-” sign before the second and third number in each of these n/4 groups. This can be understood as :
(1 – 2 – 3 + 4) + (5 – 6 – 7 + 8) + ———— + ((n – 3) – (n – 2) – (n – 1) + n) = 0. ———–(1)
4) Now considering the case when n + 1 is divisible by 4, then n = 4t – 1 , which implies n = 3 + 4(t – 1), and we can exploit the same idea by first taking care of the first three numbers as follows:
(1 + 2 – 3) + (4 – 5 – 6 + 7) + ———— + ((n – 3) – (n – 2) – (n – 1) + n) = 0. ———–(2)
So to summarize , the problem will have the following solutions :
Compute n mod 4 i.e, the remainder of the division of n by 4.
Solution 1 : If the remainder is equal to 0, insert the “+” and “-” signs as indicated by formula (1).
Solution 2 : If the remainder is equal to 3, insert the “+” and “-” signs as indicated by formula (2).
Otherwise, return the message “No solution exists for the given problem “.
- Puzzle 51| Cheryl’s Birthday Puzzle and Solution
- Puzzle 15 | (Camel and Banana Puzzle)
- Puzzle | 3 Priests and 3 devils Puzzle
- Puzzle 85 | Chain Link Puzzle
- Puzzle 81 | 100 people in a circle with gun puzzle
- Puzzle 34 | (Prisoner and Policeman Puzzle)
- Puzzle 24 | (10 Coins Puzzle)
- Puzzle | Elevator Puzzle
- Puzzle 28 | (Newspaper Puzzle)
- Puzzle 29 | (Car Wheel Puzzle)
- Puzzle 39 | (100 coins puzzle)
- Puzzle 27 | (Hourglasses Puzzle)
- Puzzle 36 | (Matchstick Puzzle)
- Puzzle 31 | (Minimum cut Puzzle)
- Puzzle 33 | ( Rs 500 Note Puzzle )
- Puzzle 38 | (Tic Tac Toe Puzzle)
- Puzzle | Words without 'a'
- Puzzle 44 | Girl or Boy
- Puzzle | Truth and Lie
- Puzzle 65 | Hats Off
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to firstname.lastname@example.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.