If there are n children then following equation can be formed.
[(216/n)-3] = [216/(n+6)]

This will give equation which can be solved to get n=18

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Given:
\(\frac{216}{n} = c\) and \(\frac{216}{n+6} = c - 3\)

Number properties solution:
1) 216 prime factored is 2^3 * 3^3, so our AC must have only factors of 2 and 3 (eliminate C)

2) Add n + 6:
A) 36 + 6 = 42
B) 24 + 6 = 30
D) 18 + 6 = 24
E) 12 + 6 = 18

n+6 must also have only factors of 2 and 3 to divide 216 evenly (eliminate A, B)

3) Notice in E) when n = 12, remaining factors are 2*3^2 = 18 (this is the number c and also n+6), HOWEVER 18 - 3 = 15, not 12, so E is out

4) Check D) if you have time:
n = 18 is 2*(3^2), remaining factors are (2^2)*3, c = 12
n+6 = 24, 3*(2^3), remaining factors are 3^2, c - 3 = 9, so 3^2, matches.

Might be easier to just check 216/9 = 24?

Algebra solution:
\(\frac{216}{n} = c\) and \(\frac{216}{n+6} = c - 3\)

1) \(\frac{216}{n+6} = \frac{216}{n - 3}\)

2) \(\frac{216}{n+6} = \frac{216 - 3n}{n}\)

3) \(216n = 216n + 6*216 - 3(n^{2}+6n)\)

4) \(0 = 6*216 - 3(n^{2}+6n)\)

5) \(3(n^{2}+6n) = 6*216\)

6) \(n^{2}+6n = 2*216\)

7) \(n^{2}+6n - 432 = 0\)

8) Factor, notice we ignore (n + root1) because n must be positive, also root 1 > root 2 because r1+r2=6 (positive) and r1*r2 = -432 (negative)

9) 432 prime factored --> (4^4)(3^3), notice 6 and -432 are even so we must have 2 even factors (multiples of 2 in each)

10) Try root 1 = 4*9 --> 36, root 2 = 4*3 --> 12, 36-12 = 24, too big, remove factor of 3 and add factor of 2 to first root

11) r1 = 8*3 --> 24, r2 = 9*2 --> 18, 24-18 = 6, so (n + 24)(n - 18) = 0

n = 18