A summer camp has 54 boys and 48 girls. They are split into groups so that every group has the same number of boys and the same number of girls, and everybody is in a group. What is the largest possible number of groups?

Step-by-step solution

1. 1

   Translate the conditions into divisibility. If there are $g$ groups with $b$ boys each and $h$ girls each, then $gb=54$ and $gh=48$. So $g$ must divide both $54$ and $48$ — the group count is a common divisor, and "as many groups as possible" means the greatest common divisor.

2. 2

   Factor both totals into primes.

   $54=2\times 3^3,\qquad 48=2^4\times 3$

3. 3

   Take the smaller exponent of each shared prime. The primes common to both are $2$ and $3$:

   $\gcd(54,48)=2^{\min(1,4)}\times 3^{\min(3,1)}=2\times 3=6$

4. 4

   Confirm the split actually works. With $g=6$,

   $\frac{54}{6}=9\ \text{boys per group},\qquad \frac{48}{6}=8\ \text{girls per group}$

   Both are whole numbers and $6\times 9=54$, $6\times 8=48$, so nobody is left over.

5. 5

   Check that $7$ or more groups is impossible. The divisors of $54$ are $1,2,3,6,9,18,27,54$ and the divisors of $48$ are $1,2,3,4,6,8,12,16,24,48$. Their common divisors are only $1,2,3,6$, so $6$ is genuinely the largest — any count above it fails to divide at least one of the two totals. As a cross-check, the Euclidean algorithm gives $54=48+6$, $48=8\times 6+0$, so $\gcd=6$.