Intuition The big idea in one breathEvery whole number bigger than 1 is like a word built from a fixed alphabet of primes . Numbers such as 4 , 6 , 12 4, 6, 12 4 , 6 , 12 broken apart into smaller factors, but primes (2 , 3 , 5 , 7 , … 2,3,5,7,\dots 2 , 3 , 5 , 7 , … atoms of multiplication. Prime factorization is the act of smashing a number down until only these atoms remain.
Intuition Why care about primes at all?
Uniqueness gives us a fingerprint. The Fundamental Theorem of Arithmetic says every number has exactly one prime factorization (ignoring order). So 12 = 2 2 ⋅ 3 12 = 2^2\cdot 3 12 = 2 2 ⋅ 3 nothing else — this fingerprint lets us compute HCF/GCD , LCM , simplify fractions, check divisibility, and count divisors.80/20 payoff: master one skill (breaking a number into primes) and you unlock a whole chapter — LCM, HCF, rationalising, cryptography roots — for free.
A prime is a natural number > 1 > 1 > 1 1 1 1 2 , 3 , 5 , 7 , 11 , 13 2,3,5,7,11,13 2 , 3 , 5 , 7 , 11 , 13 1 1 1 not prime (it has only one divisor).
Definition Prime factorizationWriting a number n > 1 n>1 n > 1 product of primes :
n = p 1 a 1 p 2 a 2 ⋯ p k a k n = p_1^{a_1}\,p_2^{a_2}\cdots p_k^{a_k} n = p 1 a 1 p 2 a 2 ⋯ p k a k p i p_i p i a i ≥ 1 a_i \ge 1 a i ≥ 1 unique up to reordering.
Definition Fundamental Theorem of Arithmetic (FTA)Every integer n > 1 n>1 n > 1 unique except for the order of the factors.
tree ?
You keep splitting a number into any two factors. Composite branches split again; prime branches stop (become leaves). The leaves collected are your primes. Different choices of split give the same leaves — that's FTA in action, and a great self-check.
Worked example Factor tree of60 60 60 60 = 6 × 10 60 = 6 \times 10 60 = 6 × 10
6 = 2 × 3 6 = 2 \times 3 6 = 2 × 3 Why this step? 2 , 3 2,3 2 , 3 10 = 2 × 5 10 = 2 \times 5 10 = 2 × 5 Collect leaves: 2 , 3 , 2 , 5 2, 3, 2, 5 2 , 3 , 2 , 5 60 = 2 × 2 × 3 × 5 = 2 2 ⋅ 3 ⋅ 5 60 = 2\times2\times3\times5 = 2^2\cdot 3\cdot 5 60 = 2 × 2 × 3 × 5 = 2 2 ⋅ 3 ⋅ 5
Steel-man check: start differently, 60 = 4 × 15 60 = 4 \times 15 60 = 4 × 15
4 = 2 × 2 4 = 2\times2 4 = 2 × 2 15 = 3 × 5 15 = 3\times5 15 = 3 × 5 2 , 2 , 3 , 5 2,2,3,5 2 , 2 , 3 , 5 Same answer. ✅ (This proves uniqueness to your own eyes.)
ladder ?
Instead of splitting into any factors, you divide out the smallest prime that fits , again and again, writing quotients down like rungs of a ladder. It's systematic — no guessing — so it's the safer method under exam pressure.
Worked example Ladder for60 60 60 smallest prime first: 2 2 2 3 3 3 5 5 5 7 … 7\dots 7 …
2 | 60
2 | 30
3 | 15
5 | 5
| 1 ← stop when you reach 1
Why start with 2? 60 60 60 2 2 2 2 2 2 Why move to 3? 15 15 15 2 2 2 15 15 15 3 3 3 Why stop at 1? When the quotient is 1 1 1 Primes down the left side: 2 , 2 , 3 , 5 2,2,3,5 2 , 2 , 3 , 5 60 = 2 2 ⋅ 3 ⋅ 5 60 = 2^2\cdot 3\cdot 5 60 = 2 2 ⋅ 3 ⋅ 5
Worked example Ladder for84 84 84
2 | 84
2 | 42
3 | 21
7 | 7
| 1
84 84 84 42 42 42 Why? even numbers always take a factor of 2 2 2 42 42 42 21 21 21 21 21 21 21 = 3 × 7 21 = 3\times 7 21 = 3 × 7 7 7 7 1 1 1 84 = 2 2 ⋅ 3 ⋅ 7 84 = 2^2\cdot 3\cdot 7 84 = 2 2 ⋅ 3 ⋅ 7
Recall Predictbefore peeking
Q: Factorize 360 360 360 2 , 3 , 5 2,3,5 2 , 3 , 5 Verify (ladder):
2 | 360 2 | 180 2 | 90 3 | 45 3 | 15 5 | 5 | 1
360 = 2 3 ⋅ 3 2 ⋅ 5 360 = 2^3\cdot 3^2\cdot 5 360 = 2 3 ⋅ 3 2 ⋅ 5 = ( 3 + 1 ) ( 2 + 1 ) ( 1 + 1 ) = 24 = (3{+}1)(2{+}1)(1{+}1)=24 = ( 3 + 1 ) ( 2 + 1 ) ( 1 + 1 ) = 24 2 3 ⋅ 3 2 ⋅ 5 2^3\cdot3^2\cdot5 2 3 ⋅ 3 2 ⋅ 5 2 2 2^2 2 2 360 360 360 360 → 180 → 90 360\to180\to90 360 → 180 → 90
Common mistake "1 is a prime, so I'll include it."Why it feels right: 1 1 1 The fix: a prime needs exactly two distinct divisors; 1 1 1 1 1 1 6 = 2 ⋅ 3 = 1 ⋅ 2 ⋅ 3 = 1 ⋅ 1 ⋅ 2 ⋅ 3 … 6 = 2\cdot3 = 1\cdot2\cdot3 = 1\cdot1\cdot2\cdot3\dots 6 = 2 ⋅ 3 = 1 ⋅ 2 ⋅ 3 = 1 ⋅ 1 ⋅ 2 ⋅ 3 … 1 1 1 unit , not a prime.
Common mistake Forgetting to write repeated primes as powers.Why it feels right: 60 = 2 ⋅ 2 ⋅ 3 ⋅ 5 60 = 2\cdot2\cdot3\cdot5 60 = 2 ⋅ 2 ⋅ 3 ⋅ 5 is correct. The fix: it's not wrong , but compress to 2 2 ⋅ 3 ⋅ 5 2^2\cdot3\cdot5 2 2 ⋅ 3 ⋅ 5 2 ⋅ 3 ⋅ 5 = 30 2\cdot3\cdot5=30 2 ⋅ 3 ⋅ 5 = 30
Common mistake Stopping the ladder too early.Why it feels right: you reach an odd number and feel done. The fix: you're done only when the quotient is 1 1 1 77 77 77 3 , 5 , 7 , 11 … 3,5,7,11\dots 3 , 5 , 7 , 11 … 77 = 7 × 11 77 = 7\times11 77 = 7 × 11
Common mistake Testing a non-prime divisor.Why it feels right: "60 ÷ 6 = 10 60 \div 6 = 10 60 ÷ 6 = 10 The fix: the ladder must use primes only , else you don't get atoms. Splitting by 6 6 6 tree (you split 6 6 6
"2-3-5-7, then knock on 11's door"
Test primes in order 2 , 3 , 5 , 7 , 11 , 13 2,3,5,7,11,13 2 , 3 , 5 , 7 , 11 , 13 squared exceeds the remaining number (then whatever's left is itself prime). E.g. for 77 77 77 7 7 7 8 2 = 64 < 77 8^2=64<77 8 2 = 64 < 77 9 2 = 81 > 77 9^2=81>77 9 2 = 81 > 77 77 ≈ 8.7 \sqrt{77}\approx8.7 77 ≈ 8.7 7 7 7 77 = 7 ⋅ 11 77=7\cdot11 77 = 7 ⋅ 11
Recall Explain simply (hidden)Imagine every number is a LEGO castle. Some blocks can be pulled into smaller blocks, but prime blocks (2 , 3 , 5 , 7 … 2,3,5,7\dots 2 , 3 , 5 , 7 … how you dismantle it (tree method) or carefully step by step (ladder method), you always end up with exactly the same pile of tiny blocks . That "same pile every time" rule is what makes math so tidy.
Why is 1 not a prime number? A prime must have exactly two distinct divisors (1 and itself); 1 has only one divisor, so it's a unit, not a prime.
State the Fundamental Theorem of Arithmetic. Every integer >1 is a product of primes, and that factorization is unique except for the order of factors.
Prime-factorize 60. 2 2 ⋅ 3 ⋅ 5 2^2\cdot3\cdot5 2 2 ⋅ 3 ⋅ 5 .
Prime-factorize 84. 2 2 ⋅ 3 ⋅ 7 2^2\cdot3\cdot7 2 2 ⋅ 3 ⋅ 7 .
In the ladder method, when do you stop? When the quotient becomes 1.
In a factor tree, which nodes are the leaves? The prime numbers (they can't be split further).
Formula for number of divisors of n = p 1 a 1 ⋯ p k a k n=p_1^{a_1}\cdots p_k^{a_k} n = p 1 a 1 ⋯ p k a k d ( n ) = ( a 1 + 1 ) ( a 2 + 1 ) ⋯ ( a k + 1 ) d(n)=(a_1+1)(a_2+1)\cdots(a_k+1) d ( n ) = ( a 1 + 1 ) ( a 2 + 1 ) ⋯ ( a k + 1 ) .
How do you get HCF and LCM from prime powers? HCF = product of primes to the min exponent; LCM = product to the max exponent.
Compute HCF and LCM of 60 and 84 via primes. HCF
= 2 2 ⋅ 3 = 12 =2^2\cdot3=12 = 2 2 ⋅ 3 = 12 ; LCM
= 2 2 ⋅ 3 ⋅ 5 ⋅ 7 = 420 =2^2\cdot3\cdot5\cdot7=420 = 2 2 ⋅ 3 ⋅ 5 ⋅ 7 = 420 .
Which prime should you always test first in the ladder? The smallest, 2 (keep dividing while even), then 3, 5, 7...
Why do different factor trees give the same answer? Because of unique factorization (FTA) — the multiset of prime leaves is fixed.
How many divisors does 360 = 2 3 ⋅ 3 2 ⋅ 5 360=2^3\cdot3^2\cdot5 360 = 2 3 ⋅ 3 2 ⋅ 5 ( 3 + 1 ) ( 2 + 1 ) ( 1 + 1 ) = 24 (3+1)(2+1)(1+1)=24 ( 3 + 1 ) ( 2 + 1 ) ( 1 + 1 ) = 24 .
HCF and LCM — computed directly from these prime powers (min/max exponents).Divisibility Rules — decide which prime to divide by next in the ladder.Fundamental Theorem of Arithmetic — guarantees the answer is unique.Simplifying Fractions — cancel shared prime factors of numerator & denominator.Number of Divisors — uses the exponent-plus-one formula.Sieve of Eratosthenes — how to generate the prime "alphabet" you test with.
Fundamental Theorem of Arithmetic
n as product of prime powers
HCF LCM divisibility fractions
Intuition Hinglish mein samjhoDekho, har number (jo 1 se bada hai) ek building ki tarah hai, aur prime numbers (2 , 3 , 5 , 7 , 11 … 2,3,5,7,11\dots 2 , 3 , 5 , 7 , 11 … Fundamental Theorem of Arithmetic kehta hai ki chahe kaise bhi todo, aakhir mein wahi primes milenge. Isliye 60 60 60 2 2 ⋅ 3 ⋅ 5 2^2\cdot3\cdot5 2 2 ⋅ 3 ⋅ 5
Do methods hain. Factor tree mein number ko kisi bhi do factors mein tod do, phir jo composite ho use aur todo, prime aayi to ruk jao (leaf ban gayi). Sab leaves collect karo — bas ho gaya. Ladder method zyada safe hai exam ke liye: sabse chota prime lo jo divide kare — pehle 2 2 2 3 3 3 5 5 5 1 1 1
Ye skill 80/20 wala hai — ek baar aa gaya to HCF (chote exponent lo, yaani min), LCM (bada exponent lo, yaani max), fraction simplify karna, aur divisor count (( a 1 + 1 ) ( a 2 + 1 ) … (a_1+1)(a_2+1)\dots ( a 1 + 1 ) ( a 2 + 1 ) … 1 1 1 1 1 1 2 2 2 3 3 3
