Question: Express $3780$ as a product of its prime factors, giving your answer in index form.

Step 1. Use the ladder method — keep dividing by the smallest prime that works:

$945$ is odd, so move up to the next prime:

Step 2. $35$ has no factor of $3$; next primes:

Reaching $1$ means the ladder is finished.

Step 3. Collect the divisors used — two $2$s, three $3$s, one $5$, one $7$ — and write in index form:

Step 4. Check by multiplying back: $4 \times 27 = 108$, $\;108 \times 5 = 540$, $\;540 \times 7 = 3780$. ✓

⚠ Watch out: Every number on the ladder's left must be prime — dividing by $6$ or $9$ breaks the method. Always finish in index form ($2^2 \times 3^3 \times 5 \times 7$, not $2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 7$); the indices are what HCF, LCM and root questions run on.

Question: Evaluate $\sqrt{144}$ using prime factorisation.

Step 1. Factorise $144$ with the ladder method:

Step 2. Square-rooting a product square-roots each factor — and for powers, divide the index by 2:

Step 3. Evaluate:

Step 4. The same idea handles cube roots — divide every index by $3$ instead. For example $\sqrt[3]{2^6 \times 5^3} = 2^2 \times 5 = 20$.

Check. $12^2 = 144$. ✓ And the method reveals why it worked: every index in $2^4 \times 3^2$ was even, which is exactly what makes $144$ a perfect square.

⚠ Watch out: Indices are divided by 2, never square-rooted themselves — $\sqrt{2^4}$ is $2^2$, not $2^{\sqrt{4}}$ by luck. If some index is odd, the number is not a perfect square and the root won't come out whole — that observation is the bridge to "smallest $k$ so that $nk$ is a perfect square" questions.

Question: Find the smallest whole number $k$ such that $60k$ is a perfect square.

Step 1. Factorise first — perfect-square questions are always answered through the indices:

Step 2. A perfect square needs every index even. Scan the indices:

$2^2$ — even ✓ $\qquad 3^1$ — odd ✗ $\qquad 5^1$ — odd ✗

Step 3. Top up each odd index by exactly one more factor:

Step 4. Confirm:

The same logic with "every index a multiple of 3" answers perfect-cube versions: $60k$ a perfect cube needs $k = 2 \times 3^2 \times 5^2 = 450$.

The move is always the same: factorise, scan the indices, buy just enough factors to fix the odd ones.

⚠ Watch out: "Smallest" means top up only what's missing — multiplying by $60$ itself gives a perfect square ($3600$) but not the smallest one. Read whether the question wants a square (even indices) or a cube (indices divisible by 3); the top-up differs.