How did the solution assume this?

• Total Sum ($S_n$): 810 (Total children)
• Common Difference ($d$): 3 (Since each row has 3 more or fewer than the next)
• Number of Rows ($n$): $N$ (This is what we need to find)
• First term ($a$): The number of children in the smallest row.

The formula for the sum of an AP is:

$S_n = \frac{n}{2} [2a + (n-1)d]$

$810 = \frac{N}{2} [2a + (N-1)3]$

$1620 = N [2a + 3(N-1)]$

$2a + 3(N-1) = \frac{1620}{N}$

$2a = \frac{1620}{N} - 3(N-1)$

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Test the Options:

For a solution to be valid, $a$ must be a positive integer. This means $2a$ must be an even integer.

Test $N = 6$:

$2a = \frac{1620}{6} - 3(5) = 270 - 15 = 255$

Since 255 is odd, $a$ would be $127.5$ (Not an integer). Invalid.

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Test $N = 8$:

$2a = \frac{1620}{8} - 3(7) = 202.5 - 21 = 181.5$

Not an integer. Invalid.

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Test $N = 10$:

$2a = \frac{1620}{10} - 3(9) = 162 - 27 = 135$

Since 135 is odd, $a$ would be $67.5$. Invalid.

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Test $N = 5$:

$2a = \frac{1620}{5} - 3(4) = 324 - 12 = 312$

$a = \frac{312}{2} = 156$

Since 156 is a whole number, this works! Correct.

Conclusion: Only $N = 5$ allows for a whole number of children in each row.