The HCF of two whole numbers is 12, and the two numbers add up to 84. How many such pairs are there?

Write the numbers as $12a$ and $12b$, where a and $b$ share no common factor. Then $12(a+b)=84$, so $a+b=7$. The ways to split 7 are 1 + 6, 2 + 5 and 3 + 4, and every one of these pairs is co-prime, so all three count: the numbers are (12, 72), (24, 60) and (36, 48). Each pair really does have HCF 12. Counting the ordered versions as separate — (12, 72) and (72, 12) — doubles the answer to 6, which is the trap here.