Every whole number is a product of primes in exactly one way. Finding it opens up half the number-sense questions on the paper.
The problem
Write 360 as a product of primes.
360 = 36 × 10 = (4 × 9)(2 × 5) = 2² × 3² × 2 × 5 = 2³ × 3² × 5. B is not a prime factorisation at all — 8 and 45 are composite. Multiply back to check: 8 × 9 × 5 = 360. ✓
360 ÷ 2 ÷ 2 ÷ 2 = 45
take out every 2 first
45 ÷ 3 ÷ 3 = 5
2 no longer fits, move to 3
5 ÷ 5 = 1
stop at 1
2³ · 3² · 5
count the repeats
Starting elsewhere is fine — 360 = 36 × 10 breaks down to the same thing. The uniqueness is guaranteed, so any route that ends in primes ends in these primes.
| Divisible by | When | 360 |
|---|---|---|
| 2 | last digit is even | yes |
| 3 | digit sum is divisible by 3 | 3 + 6 + 0 = 9, yes |
| 4 | last two digits divisible by 4 | 60, yes |
| 5 | ends in 0 or 5 | yes |
| 6 | divisible by 2 and by 3 | yes |
| 8 | last three digits divisible by 8 | 360, yes |
| 9 | digit sum divisible by 9 | 9, yes |
| 11 | alternating digit sum divisible by 11 | 3 − 6 + 0 = −3, no |
There is no simple rule for 7 — just divide and see. That is why 7 is where the work starts on awkward numbers.
A factor above √n would need a partner below it — and you have already checked those.
√391 ≈ 19.8
so test primes up to 19
2, 3, 5 — no
odd, digit sum 13, does not end in 0 or 5
7, 11, 13 — no
391 ÷ 7 = 55.9, ÷ 11 = 35.5, ÷ 13 = 30.1
17: 391 ÷ 17 = 23
a hit
391 = 17 × 23
not prime
Factors come in pairs multiplying to n. If both were bigger than √n their product would exceed n, so every composite number has a factor at or below its square root. Seven primes settle 391; without the rule you would try 200 numbers.
| Number | Prime? | Why |
|---|---|---|
| 1 | no | it has only one divisor; a prime needs exactly two |
| 2 | yes | the only even prime |
| 9, 15, 21 | no | odd is not the same as prime |
| 51 | no | 3 × 17 — digit sum 6 gives it away |
| 91 | no | 7 × 13, the classic “looks prime” |
| two primes adding to 39 | 2 and 37 | an odd sum needs one even prime, and 2 is the only one |
| 97 | yes | nothing up to 9 divides it |
Primes under 50, worth knowing cold: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.
A large prime at the end: factorise 354.
354 ÷ 2 = 177
even
177 ÷ 3 = 59
digit sum 15
59: test 2, 3, 5, 7 — none divide
√59 ≈ 7.7, so 59 is prime
354 = 2 × 3 × 59
the ladder ends at a prime, not at 1
What is different: the last factor is large. Once the leftover passes the square-root test, we stop — there is nothing more to find.
A power, without multiplying it out: write 12⁴ as a product of primes.
12 = 2² × 3
factorise the base only
12⁴ = (2²)⁴ × 3⁴ = 2⁸ × 3⁴
multiply each exponent by 4
What is different: 12⁴ = 20,736 is slow to break down. The power multiplies the exponents — 2 × 4 = 8, not 2 + 4 = 6.
A perfect cube instead of a square: find the smallest k that makes 72k a perfect cube.
72 = 2³ × 3²
exponents 3 and 2
a cube needs multiples of 3: 3² needs one more 3
2³ is already fine
k = 3, and 72 × 3 = 216 = 6³
check ✓
What is different: for a cube we round each exponent up to the next multiple of 3. Thinking “square” gives k = 2, and 144 is not a cube.
| Task | Using 360 = 2³ · 3² · 5 |
|---|---|
| GCF and LCM | lowest powers for GCF, highest for LCM |
| counting divisors | (3+1)(2+1)(1+1) = 24 of them |
| simplifying √360 | √2²·3²·10 = 6√10 |
| is it a perfect square? | no — the 5 has an odd exponent |
A perfect square has every exponent even; a perfect cube has every exponent a multiple of three. So 360 needs one more 2 and one more 5 to become a square: 360 × 10 = 3600 = 60².
Expect “is it prime” (test primes only up to the square root), a product of three primes to unpick, two primes with an odd sum (one of them is 2), and the smallest k that makes a number a perfect square. Every one of these starts with the factor ladder — write it first.
Is it prime?
Is 247 prime?
Test primes up to √247 ≈ 15.7 — that is 2, 3, 5, 7, 11, 13. It survives until 13, and 13 × 19 = 247. B and C are not even correct arithmetic (7 × 35 = 245, 3 × 82 = 246). Stopping at the square root is what keeps this quick.
