A group of 630 children are arranged in rows for a group photo session. Each…
2022
A group of 630 children are arranged in rows for a group photo session. Each row contains three less children than the row in front of it. What number of rows is not possible?
- A.
3
- B.
4
- C.
5
- D.
6
Attempted by 53 students.
Show answer & explanation
Correct answer: D
Key idea: Let n be the number of rows and a be the number of children in the first row. The rows form an arithmetic sequence a, a-3, a-6, ..., a-3(n-1).
Sum formula:
Total children 630 = n/2 · (first term + last term) = n/2 · (a + (a - 3(n-1))) = n/2 · (2a - 3(n-1)).
Rearrange to solve for a: 2a - 3(n-1) = 1260/n, so a = (1260/n + 3(n-1))/2.
Requirements for a valid arrangement:
1) a must be a positive integer.
2) The last row a - 3(n-1) must be at least 1 (positive).
Check each candidate number of rows:
3 rows: a = (1260/3 + 3·2)/2 = (420 + 6)/2 = 213. Last row = 213 - 6 = 207 > 0. Valid.
4 rows: a = (1260/4 + 3·3)/2 = (315 + 9)/2 = 162. Last row = 162 - 9 = 153 > 0. Valid.
5 rows: a = (1260/5 + 3·4)/2 = (252 + 12)/2 = 132. Last row = 132 - 12 = 120 > 0. Valid.
6 rows: a = (1260/6 + 3·5)/2 = (210 + 15)/2 = 225/2 = 112.5, which is not an integer. Since the first row must have an integer number of children, 6 rows is not possible.
Conclusion: The arrangement with 6 rows is not possible because it requires a non-integer first-row count; the other checked options are possible.