Given: Total children = 630 Let x = number of children in the first row Each subsequent row has 3 fewer children than the previous row Number of rows = n

Formula for sum of an arithmetic series: $S_n = (n/2) \cdot (2x + (n-1) \cdot (-3))$

Checking possible values of n:

For n = 3: $(3/2) \cdot (2x - 6) = 630$ $3 \cdot (2x - 6) = 1260$ $6x - 18 = 1260$ $6x = 1278$ $x = 213$

For n = 4: $(4/2) \cdot (2x - 9) = 630$ $2 \cdot (2x - 9) = 630$ $4x - 18 = 630$ $4x = 648$ $x = 162$

For n = 5: $(5/2) \cdot (2x - 12) = 630$ $5 \cdot (2x - 12) = 1260$ $10x - 60 = 1260$ $10x = 1320$ $x = 132$

For n = 6: $(6/2) \cdot (2x - 15) = 630$ $3 \cdot (2x - 15) = 630$ $6x - 45 = 630$ $6x = 675$ $x = 112.5$ (not an integer)

Conclusion: 6 rows are not possible.