How many children does Brenda's family actually have? (I)

It is natural to consider the possibility that one or more of children have their birthdays during the 9-month period but we can prove that this is impossible.

Let the required number of children (including Billy) be $x$ . Before Billy's birth, the total age is $20(x-1)$ . Assuming $y$ out of $(x-1)$ siblings have their birthdays during this period, the total age would go up by exactly $y$ years as at most one birthday can possibly happen during a period of 9 months.

This means that when Billy was born, the total age of the $x$ children including Billy would be $20(x-1) + y$ . Since we also know that the average age of these $x$ children is 16 years, making the total age after Billy's birth $16x$ , we have $20(x-1)+y = 16x$ or $y=20-4x$ . Note that $20-4x$ is divisible by 4 so $y$ should likewise be a multiple of 4. Furthermore, it only makes sense that $0 </=y </= (x-1)$ . If $y <0$ , it implies that the children became younger which is absurd. Moreover, we are looking at $y$ out of $(x-1)$ children. hence $y$ cannot possibly exceed $(x-1)$ .

Plugging this inequality into the equation $x =\frac{20-y}{4}$ to get $y </=\frac{20-y}{4}-1$ , what we would end up with is $5y </= 16$ . Bearing in mind that $y$ is a non-negative integer divisble by 4, the only possible value is 0. In other words, it is not at all possible that one of the siblings had their birthdays during the 9-month period.

Alternatively, knowing that $x =\frac{20-y}{4}$ , if we were to try the smallest positive multiple of 4 (i.e. $y=4$ ), we have $x = \frac{20-4}{4} = \frac{16}{4} = 4$ but we realise that $y = 4$ while $x-1 = 3$ which does not make sense as $y> x-1$ . Since $x$ is a decreasing function on $y$ , larger multiples of 4 for $y$ would result in an even smaller $x$ so it is clear that for larger multiples of 4, $y>x-1$ . as well. Moreover, since $y$ is non-negative, $y=0$ .

Since $y=0$ , we now have $20(x-1) = 16x$

$20x - 20 = 16x$

$4x = 20$

$x=\boxed{5}$