There are 240 balls and n number of boxes B1, B2, B3, ... , Bn. The balls are to be placed in the boxes such that should contain 4 balls more than B2, B2 should contain: 4 balls more than B3, and so on. Which one of the following cannot be the possible va

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Q: 75 (IAS/2009)
There are 240 balls and n number of boxes B1, B2, B3, ... , Bn. The balls are to be placed in the boxes such that should contain 4 balls more than B2, B2 should contain: 4 balls more than B3, and so on.
Which one of the following cannot be the possible value of n ?

question_subject: 

Logic/Reasoning

question_exam: 

IAS

stats: 

0,1,7,3,2,2,1

keywords: 

{'boxes b1': [0, 0, 1, 0], 'possible value': [0, 0, 2, 0], 'boxes': [0, 0, 0, 1], 'balls': [0, 1, 1, 0], 'b3': [0, 0, 1, 1], 'b2': [0, 0, 0, 1], 'number': [0, 0, 0, 2], 'bn': [0, 0, 1, 0]}

The problem involves distributing balls into a series of boxes so that each box has 4 balls more than the next one. This creates an arithmetic series. The sum of an arithmetic series can be given by the formula: n/2(2a+(n-1)d), where `n` is the number of terms (boxes), `a` is the first term, and `d` is the common difference.

Given we have 240 balls, the number of balls in the first box is 4(n-1) since each box has 4 balls more than the next one. Applying the sum formula, we get

240 = n/2 [2*4(n-1) + (n-1)4]

240 = 2n^2 - 2n

Solving this quadratic equation, we find no integer solutions for n=7. The answer options `4`, `5`, `6` yield integer solutions satisfying the condition, but `7` doesn`t. Hence, the number of boxes cannot be 7 i.e., option-4.