Fill in the blanks: 99 = ? x ? x ?

Step-by-step solution

1. 1

   Test the small primes in order. $99$ is odd, so $2$ is out. The digit sum is $9+9=18$, a multiple of $9$, so $99$ is divisible by both $3$ and $9$:

   $99\div9=11$

   giving the two-factor split $99=9\times11$. The divisibility-by-3 digit-sum test is what makes this immediate without trial division.

2. 2

   Break any composite factor further. $11$ is prime (not divisible by $2$, $3$, or any prime up to $\sqrt{11}\approx3.3$), but $9$ is not:

   $9=3\times3$

3. 3

   Assemble the three-factor product.

   $99=3\times3\times11$

   Check: $3\times3=9$ and $9\times11=99$ ✓. This is the prime factorisation, usually written $99=3^{2}\times11$.

4. 4

   Confirm it is the only genuine answer. By unique factorisation, $99$ has exactly the prime factors $3,3,11$, so any three-factor product without using $1$ must be a rearrangement of these. Writing $99=1\times9\times11$ is also technically three factors, but padding with $1$ adds nothing — the intended answer uses three factors greater than $1$.

5. 5

   Cross-check by listing all divisors. From $3^{2}\times11$ the divisors are $1,3,9,11,33,99$ — six of them, matching the count $(2+1)(1+1)=6$ predicted by the exponents. Every divisor appears in a factor pair: $1\times99$, $3\times33$, $9\times11$ ✓.