I’m reviewing prime factorisation and I keep tripping over something that feels basic. I was taught to break a number into factors until everything is prime. But I keep assuming the first split matters. For example, with 180, if I start with 18×10 instead of 12×15, I end up with a different list of primes at the end, so I’ve been thinking there can be multiple correct prime factorizations for the same number. Now I’m being told the prime factorization is unique, and I’m not seeing how that fits with what I’m getting.
Could someone explain what I’m doing wrong in my logic here? If I choose different starting pairs for the same number, why shouldn’t I get different primes at the end? Is there a rule that forces the same set of primes and exponents no matter which path I take?
I remember struggling with this in school – my factor trees for numbers like 84 and 360 rarely matched my classmates’. I chalked it up to messy branching, but I’m still running into the same issue now when I try to check my work quickly during practice tests.
As a quick check, I’ve been adding up the primes I get at the end of a factor tree and comparing that sum between different trees; if the sums match, I’ve been assuming both factorizations are correct. Is that a valid shortcut, or is there a better quick test? Also, for a number like 144, since it’s 12×12, does that mean any prime that shows up in 12 only needs to appear once overall, not twice? That feels right to me, but I’m not fully confident.
Ready to make maths more enjoyable, accessible, and fun? Join a friendly community where you can explore puzzles, ask questions, track your progress, and learn at your own pace.
By becoming a member, you unlock:
3 Responses
Great question! The key fact you’re bumping into is the Fundamental Theorem of Arithmetic: every integer greater than 1 has a unique prime factorization, up to the order of the primes. Different factor trees just reorder the same primes; if you fully split each branch down to primes and then collect like primes, you’ll always land on the same multiset. For example, 180 = 18×10 gives 18 = 2×3² and 10 = 2×5, so 180 = 2²×3²×5; starting 12×15 gives 12 = 2²×3 and 15 = 3×5, which again multiplies to 2²×3²×5. The reason this must happen is Euclid’s lemma: if a prime p divides ab, then p divides a or b; that forces the same primes to appear with the same exponents no matter how you split. What might be going wrong is stopping too early or not gathering like primes correctly; make sure each leaf is prime and then add exponents across factors. For 144, since 12 = 2²×3, we get 144 = 12×12 = (2²×3)² = 2⁴×3², so the primes from 12 appear twice, not once. As a quick check, don’t add the primes-sums can match even when products don’t (e.g., 2+2+2+3+5 = 14 and 2+2+3+7 = 14, but 2·2·2·3·5 = 120 and 2·2·3·7 = 84). A better quick test is to sort the prime lists and compare term by term, or just multiply the primes back to see if you recover the original number. Hope this helps!
Totally get the feeling-no matter how you split, it’s like rearranging the same Lego bricks: 180 = 18*10 = (2*3^2)*(2*5) -> 2^2*3^2*5, and 180 = 12*15 = (2^2*3)*(3*5) -> 2^2*3^2*5. I’ll even do a quick (kinda sloppy) “sum of the primes” check to compare trees, but what really matters-so 144 = 12*12 gives 2^4*3^2-is the exponents; nice explainer here: https://www.khanacademy.org/math/arithmetic/factors-multiples/prime_factorization/a/prime-factorization.
Yep-prime factorization is unique: different factor trees just rearrange the same multiset of primes with the same exponents, like breaking a LEGO model into its smallest bricks-no matter where you start pulling, you end up with the same bag of basic pieces. Don’t use the sum-of-primes check (it can fool you); instead multiply the primes back or compare exponents, and for 144 = 12×12 you get all the primes from 12 twice, so 144 = (2^2·3)^2 = 2^4·3^2.