Find the highest common factor of p = 3^3 x 5^4 x 7^5 x 11 and q = 3 x 5^3 x 7^4. Give your answer in index form.

Step-by-step solution

1. 1

   Notice the numbers are already factorised. Both $p$ and $q$ are handed to you as products of prime powers. That is the whole gift of this question: the hard part of an HCF — breaking each number into primes — is already done, so you must not multiply out. $p$ alone is $3^3\times5^4\times7^5\times11 = 3{,}119{,}194{,}687{,}500$, and factorising that back would be pointless work.

2. 2

   Recall the index rule for an HCF. The HCF is the largest number dividing both. A prime power $r^k$ divides $p$ only if $k$ is at most the exponent of $r$ in $p$, and divides $q$ only if $k$ is at most the exponent in $q$. So for each prime, take the smaller of the two exponents:

   $\text{HCF} = \prod_r r^{\min(a_r,\,b_r)}$

3. 3

   Line the primes up side by side. Write each number's exponent for every prime that appears, using exponent $0$ where a prime is missing:

   prime	in $p$	in $q$	smaller
   $3$	$3$	$1$	$1$
   $5$	$4$	$3$	$3$
   $7$	$5$	$4$	$4$
   $11$	$1$	$0$	$0$

4. 4

   Handle the prime 11 carefully. This is where the mark is usually lost. $11$ appears in $p$ but not in $q$, so its exponent in $q$ is $0$ and $\min(1,0)=0$. Since $11^0=1$, the factor $11$ drops out of the HCF entirely — an HCF can never contain a prime that only one of the numbers has.

5. 5

   Assemble the answer in index form.

   $\text{HCF}(p,q) = 3^1 \times 5^3 \times 7^4 = 3 \times 5^3 \times 7^4$

6. 6

   Check by division. Multiplying out gives $3\times125\times2401 = 900{,}375$. Dividing: $q \div 900{,}375 = 1{,}801{,}500 \div \ldots$ — more simply, $q = 3\times5^3\times7^4 = 900{,}375$ exactly, so the HCF equals $q$ itself, which makes sense because every exponent of $q$ is at most the matching exponent of $p$. Hence $q$ divides $p$, and $900{,}375$ divides both.