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 numberof rows is not possible

The problem can be solved by using the formula for the sum of an arithmetic series. The sum S of an arithmetic series is given by the formula S = n/2 * (a + l), where n is the number of terms, a is the first term, and l is the last term.

In this case, we know that the sum S is 630 (the total number of children), and we know that each row has 3 fewer children than the row in front of it. This means that the difference between each term in the series is -3.

We can start by assuming that the first row (the one with the most children) has x children. Then the last row (the one with the least children) would have x - 3*(n-1) children, where n is the number of rows.

Substituting these values into the formula for the sum of an arithmetic series, we get:

630 = n/2 * (2x - 3*(n-1))

Solving this equation for n, we get:

n = 1260 / (2x - 3*(n-1))

Since n must be a positive integer (you can't have a fraction of a row), the number of rows is not possible if the right-hand side of this equation is not a positive integer. This will happen if 2x - 3*(n-1) does not evenly divide 1260.

So, to find the number of rows that is not possible, you would need to try different values of x and n and see which ones do not result in n being a positive integer.