Question

A group of 630 children is arranged in rows for a group photograph. Each row contains three fewer children than the row in front of it. What number of rows is not possible?

• A. 6
• B. 3
• C. 5
• D. 4

Hint:

When solving problems involving arithmetic progressions, carefully use the formula for the sum of the terms and check for integer solutions.

Answer 1

The Correct Option is A

Let the number of rows be \( r \) and the number of children in the first row be \( x \). Each subsequent row has 3 fewer children than the row in front of it. The total number of children is the sum of the children in each row. This can be written as: \[ x + (x - 3) + (x - 6) + \dots = 630 \] This is an arithmetic progression with the first term \( x \), the common difference \( -3 \), and the sum of the first \( r \) terms. The sum of the first \( r \) terms of an arithmetic progression is given by the formula: \[ S_r = \frac{r}{2} \left( 2x - 3(r - 1) \right) \] We are told that the total number of children is 630, so: \[ \frac{r}{2} \left( 2x - 3(r - 1) \right) = 630 \] We now test each of the options for \( r \) to find which one does not result in an integer value for \( x \). - For \( r = 6 \), solving the equation gives a non-integer value for \( x \), so 6 rows is not possible. - For other values of \( r \), solving the equation gives integer values for \( x \). Thus, the answer is \( \boxed{6} \).