3. Mrs Sogiba is catering for her son's birthday party. She has invited some of her son's friends to the party. The number of friends is a composite multiple of her son's age on his birthday. She has baked 48 cupcakes and each child will receive exactly the same number of cupcakes. Use your knowledge of multiples and factors to find out how old her son will be on his birthday. (This is not his first birthday party.) Question 1 For the analogue clock to show the correct time again, it must gain exactly 12 hours. (Remember there is no difference between a.m. and p.m. on an analogue clock.) Question 2

Let \( a \) be the son’s age and let the number of friends be \( m \times a \), where \( m \) is a composite number (that is, \( m>1 \) and not prime). Since Mrs Sogiba baked 48 cupcakes and gave each child the same whole number of cupcakes, the number of friends, namely \[ m \times a, \] must be a divisor of 48. We can summarize the conditions as follows: 1. \( a > 1 \) (this isn’t his first birthday). 2. \( m \) is composite (the smallest composite number is 4). 3. \( m \times a \) is a divisor of 48. Since we are looking for plausible ages for a birthday party, we look for values \( a \) (typically a small positive integer) such that there is at least one composite \( m \) with \[ m \times a \mid 48. \] A very natural way to begin is to try the smallest composite \( m \), namely \( m=4 \). Then we require \[ 4a \mid 48. \] This implies that \( 4a \) is one of the divisors of 48. In particular, we must have \[ 4a\le48\quad\Longrightarrow\quad a\le12. \] Furthermore, since \( 4a \) divides 48, we have \[ a\mid\frac{48}{4}=12. \] Thus the possible values for \( a \) are the divisors of 12 that are greater than 1: \[ a\in \{2,3,4,6,12\}. \] In many such problems the answer is unique based on context. Considering a birthday party where the number of invited friends is a reasonable number, one common choice is \( a=6 \) (so the child is 6 years old) because then the number of friends is: \[ 4\times 6=24, \] and indeed \( 24 \) divides \( 48 \) (each friend would get \( \frac{48}{24}=2 \) cupcakes). Thus the son is 6 years old on his birthday. --- For the second question the statement is: “For the analogue clock to show the correct time again, it must gain exactly 12 hours.” On an analogue clock the display “resets” every 12 hours because the numbers 1–12 repeat. Therefore, regardless of its speed (if it is running fast), it will show the correct time again when it has gained a full 12-hour cycle. So the clock must gain \( 12 \) hours in order to show the correct time again. --- Answer Summary: - The son is \( 6 \) years old. - The clock must gain \( 12 \) hours.