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

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Solution

The correct option is B 7
Common difference $= 4$
It can be taken as AP series.
$B 2 - B 1 = B 3 - B 2........ = 4$
TOTAL NUMBER OF BALLS $= 240$
Therefore, $\frac{n}{2} (2 a + (n - 1) d) = 240$
or, $2 a - (n - 1) 4 = \frac{480}{n}$
RHS should give an integer value since it is LHS has one integer value. Ths $480$ should get divisible by n. $480$ is not divisible by $7$ , thus $7$ cannot be the possible value of n.

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