Q199
ArithmeticNumber Series & Progressions
A group of 630 children 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?
- a.3
- b.4
- c.5
- d.6
Answer: (D) 6
The rows form an arithmetic progression with common difference 3. If the front (largest) row has $a$ children and there are $n$ rows, the total is $na - \frac{3n(n-1)}{2} = 630$, so $a = \frac{630}{n} + \frac{3(n-1)}{2}$, which must be a whole number. For $n = 3$, $a = 213$; for $n = 4$, $a = 162$; for $n = 5$, $a = 132$; but for $n = 6$, $a = 105 + 7.5 = 112.5$, which is impossible. Check for $n = 5$: $132 + 129 + 126 + 123 + 120 = 630$. Shortcut: with an odd number of rows the middle row equals the average ($\frac{630}{n}$), so $n$ need only divide 630; with an even number of rows the average must end in .5, which fails for 6 ($\frac{630}{6} = 105$).