A group of 630 children is arranged in rows for a group photograph session.

Each row contains three fewer children than the row in front of it.

What number of rows is not possible?

Step 1: Set up AP
Let first row (front) have $a$ children, common difference $d = -3$

For $n$ rows, sum $= 630$:
$S_n = \dfrac{n}{2}[2a + (n-1)(-3)] = 630$

$n[2a - 3(n-1)] = 1260$

$2a = \dfrac{1260}{n} + 3(n-1)$

$a = \dfrac{630}{n} + \dfrac{3(n-1)}{2}$

Step 2: Conditions for valid solution
$a$ must be a positive integer, and last row $= a + (n-1)(-3) > 0$

$\Rightarrow a > 3(n-1)$

$\Rightarrow \dfrac{630}{n} + \dfrac{3(n-1)}{2} > 3(n-1)$

$\Rightarrow \dfrac{630}{n} > \dfrac{3(n-1)}{2}$

$\Rightarrow 1260 > 3n(n-1)$

$\Rightarrow n(n-1) < 420$

$\Rightarrow n \leq 21$ (since $21 \times 20 = 420$, not $< 420$, so $n \leq 20$)

Step 3: Also $a$ must be a positive integer
$a = \dfrac{630}{n} + \dfrac{3(n-1)}{2}$ must be a positive integer.

For $a$ to be integer: $\dfrac{630}{n}$ and $\dfrac{3(n-1)}{2}$ must together give integer.

Check $n = 6$: $a = \dfrac{630}{6} + \dfrac{3(5)}{2} = 105 + 7.5 = 112.5$ — not integer! ✗

Check $n = 7$: $a = \dfrac{630}{7} + \dfrac{3(6)}{2} = 90 + 9 = 99$ ✓
Check $n = 9$: $a = \dfrac{630}{9} + \dfrac{3(8)}{2} = 70 + 12 = 82$ ✓
Check $n = 14$: $a = \dfrac{630}{14} + \dfrac{3(13)}{2} = 45 + 19.5 = 64.5$ — not integer! ✗