Question

A group of 630 children is arranged in rows for a photograph session. Each row contains three children lesser than the row in front of it. Which of the below mentioned number of rows is not possible?

• A. 3
• B. 4
• C. 5
• D. 6
• E. 7

Answer 1

The Correct Option is D

The problem involves arranging a group of 630 children in rows such that each subsequent row has three fewer children than the row in front of it. We need to determine which number of rows is not possible. 

Let's define:

• \(n\): the total number of rows.
• \(x\): the number of children in the first row.

The sequence of children in each row forms an arithmetic sequence: \(x, (x-3), (x-6), \ldots\)

The total number of children in all rows can be expressed by the formula for the sum of an arithmetic sequence:

\(S = \frac{n}{2} \left(2x - 3(n - 1) \right)\)

In this case, \(S = 630\). Therefore, the equation becomes:

\(630 = \frac{n}{2} \left(2x - 3(n - 1) \right)\)

Rewriting and solving for \(x\) gives:

\(1260 = n(2x - 3n + 3)\)

We need to test each of the given options to determine the feasibility:

1. For 3 rows: \(1260 = 3(2x - 6 + 3) = 6x - 9\)
   Solving gives \(x = \frac{1269}{6} \approx 211.5\) (Not possible since \(x\) must be an integer).
2. For 4 rows: \(1260 = 4(2x - 12 + 3) = 8x - 36\)
   Solving gives \(x = \frac{1296}{8} = 162\) (Possible).
3. For 5 rows: \(1260 = 5(2x - 15 + 3) = 10x - 60\)
   Solving gives \(x = \frac{1320}{10} = 132\) (Possible).
4. For 6 rows: \(1260 = 6(2x - 18 + 3) = 12x - 90\)
   Solving gives \(x = \frac{1350}{12} = 112.5\) (Not possible as \(x\) must be an integer).
5. For 7 rows: \(1260 = 7(2x - 21 + 3) = 14x - 126\)
   Solving gives \(x = \frac{1386}{14} = 99\) (Possible).

Hence, the number of rows that is not possible is 6 because it does not yield an integer value for \(x\).