Question: Find the HCF and the LCM of $132$, $156$ and $180$.

Step 1. Prime-factorise all three:

Step 2. HCF — take the minimum power of each common prime. Only $2$ and $3$ appear in all three:

Step 3. LCM — take the maximum power of every prime that appears anywhere:

Step 4. Sense-check both: $12$ divides each of the three numbers ($132 \div 12 = 11$, $156 \div 12 = 13$, $180 \div 12 = 15$) ✓, and $25740$ is a multiple of each ✓.

Useful cross-check for two numbers: HCF × LCM = product of the numbers (this shortcut does not extend to three).

⚠ Watch out: The rules point in opposite directions — HCF takes the lowest power of the shared primes; LCM takes the highest power of all primes, including ones ($5$, $11$, $13$) that appear in just one number. Leaving those singletons out of the LCM is the classic three-number error.

Question: Huixian packs $171$ pens, $63$ pencils and $27$ erasers into identical gift bags so that each item is equally distributed among the bags. Find the largest number of bags she can pack, and the number of each item in one bag.

Step 1. Translate: the number of bags must divide $171$, $63$ and $27$ exactly — and we want the largest such number → HCF.

Step 2. Factorise:

Step 3. Minimum power of the common prime $3$:

Step 4. Contents of each bag — divide each count by 9:

Spotting the type: "largest number… divided equally" → HCF; "smallest number… happen together again / same amount of each" → LCM (e.g. bags of 15 and 18 ribbons → LCM 90).

⚠ Watch out: Read which quantity the question wants at the end — the number of bags (the HCF) or the number of items per bag (the counts divided by it). Answering 9 when the question asked "how many pens per bag" throws away a found answer.