A group of 630 children is seated in rows for a group photo session. Each row contains three less children than the row in front of it. Which one of the following number of rows is not possible?

Let 'a' be the number of children in the first row and 'n' be the total number of rows. The number of children in each subsequent row decreases by 3. This forms an arithmetic progression (AP) with the first term 'a' and common difference 'd = -3'. The total number of children is 630, which is the sum of the AP. The formula for the sum of an AP is S_n = n/2 * [2a + (n-1)d]. Substituting S_n = 630 and d = -3: 630 = n/2 * [2a + (n-1)(-3)] Multiply by 2: 1260 = n * [2a - 3(n-1)] For a valid arrangement, 'a' (number of children in the first row) must be a positive integer, and the number of children in the last row (a - 3(n-1)) must also be a positive integer. Let's test each option for 'n': A) If n = 3: 1260 = 3 * [2a - 3(3-1)] 1260 = 3 * [2a - 6] 420 = 2a - 6 426 = 2a a = 213. (This is an integer. Last row: 213 - 3(2) = 207 > 0. Possible) B) If n = 4: 1260 = 4 * [2a - 3(4-1)] 1260 = 4 * [2a - 9] 315 = 2a - 9 324 = 2a a = 162. (This is an integer. Last row: 162 - 3(3) = 153 > 0. Possible) C) If n = 5: 1260 = 5 * [2a - 3(5-1)] 1260 = 5 * [2a - 12] 252 = 2a - 12 264 = 2a a = 132. (This is an integer. Last row: 132 - 3(4) = 120 > 0. Possible) D) If n = 6: 1260 = 6 * [2a - 3(6-1)] 1260 = 6 * [2a - 15] 210 = 2a - 15 225 = 2a a = 225 / 2 = 112.5. (This is not an integer.) Since the number of children in a row must be an integer, 'a' cannot be 112.5. Therefore, 6 rows is not a possible arrangement. The final answer is D