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 correct answer is option (D):

6

Let's analyze the problem. We have a total of 630 children arranged in rows, and each row has three fewer children than the row in front of it. This means the number of children in each row forms an arithmetic sequence.

Let n be the number of rows.
Let a be the number of children in the first row.
Let d be the common difference between the rows (which is −3, since each row has 3 fewer children than the previous).

The sum of an arithmetic series is given by: S = (n/2) · [2a + (n−1)d]

In this case, S = 630, and d = −3. So we have:

630 = (n/2) · [2a + (n−1)(−3)]
1260 = n · [2a − 3n + 3]

Now we need to check each of the given options for n (number of rows) to see if we can find a valid integer value for a (the number of children in the first row). The crucial point is that a must be a positive integer, as it represents the number of children in a row.

Let's test each option:

• n = 3:
  1260 = 3 · (2a − 9 + 3) ⇒ 1260 = 3 · (2a − 6) ⇒ 420 = 2a − 6 ⇒ 426 = 2a ⇒ a = 213. This is a valid solution.
• n = 4:
  1260 = 4 · (2a − 12 + 3) ⇒ 1260 = 4 · (2a − 9) ⇒ 315 = 2a − 9 ⇒ 324 = 2a ⇒ a = 162. This is a valid solution.
• n = 5:
  1260 = 5 · (2a − 15 + 3) ⇒ 1260 = 5 · (2a − 12) ⇒ 252 = 2a − 12 ⇒ 264 = 2a ⇒ a = 132. This is a valid solution.
• n = 6:
  1260 = 6 · (2a − 18 + 3) ⇒ 1260 = 6 · (2a − 15) ⇒ 210 = 2a − 15 ⇒ 225 = 2a ⇒ a = 112.5. This is not a valid solution because a must be an integer, and here it is not.
• n = 7:
  1260 = 7 · (2a − 21 + 3) ⇒ 1260 = 7 · (2a − 18) ⇒ 180 = 2a − 18 ⇒ 198 = 2a ⇒ a = 99. This is a valid solution.

Therefore, the only option that doesn't yield a valid, integer value for a is when the number of rows, n, is 6. This is because we can't have a fraction of a child.