Number System — Factors, Divisibility & Properties

Number System is a high-yield CAT QA chapter — a reliable 2–4 questions per paper, and the foundation for remainders, cyclicity, and much of Modern Math. Master factors, divisibility and prime factorisation and you unlock a big slice of QA.

Prime factorisation is the master key

Write N = p^a · q^b · r^c (primes). Then everything follows:

Quantity	Formula	For N = 12 = 2²·3¹
Number of factors	(a+1)(b+1)(c+1)	3·2 = 6
Sum of factors	Π (p^(k+1) − 1)/(p − 1)	(7)(4) = 28
Product of factors	N^(d/2), d = #factors	12³ = 1728

(Factors of 12: 1, 2, 3, 4, 6, 12 — six of them, summing to 28, product 1728. ✓)

Divisibility rules (memorise)

• 2/4/8: last 1/2/3 digits divisible by 2/4/8.
• 3/9: digit sum divisible by 3/9.
• 11: (sum of odd-position digits) − (sum of even-position) divisible by 11.
• 6/12: divisible by both co-prime factors (2 and 3; 3 and 4).

Method for "how many factors / divisors"

1. Prime-factorise N.
2. Apply (a+1)(b+1)... for the count.
3. For "even factors", "factors divisible by k", adjust the exponent ranges.

Exam Tricks & Tips

• 🎯 Number of factors = product of (exponent + 1). 72 = 2³·3² → 4·3 = 12 factors.
• 🎯 Product of all factors = N^(d/2) — instantly gives huge products without listing.
• 🎯 Perfect squares have an ODD number of factors (one factor √N is unpaired) — a fast classification trick.
• 🎯 Number of factors is odd ⟺ N is a perfect square — use it both directions.
• 🎯 Composite divisibility via co-primes: to test ÷12, check ÷3 and ÷4 (co-prime); don't check ÷2 and ÷6 (not co-prime).
• ❌ Common mistake: forgetting to add 1 to each exponent — the count is (a+1)(b+1)..., not a·b·c.

Expected exam pattern

2–4 QA questions on factor counts, divisibility conditions, digit problems, and "how many numbers below N satisfy…". MCQ + TITA. Often combined with remainders or last-digit logic. Foundational — pays across the whole QA section.

Quick recap

Prime-factorise first; then factor count = Π(exponent+1), product of factors = N^(d/2). Perfect squares have an odd factor count. Test composite divisibility through co-prime factors, and never forget the "+1" in the count formula.

Cover the answer, recall, then check. 12 cards on factors and divisibility.

Q1. N = p^a·q^b·r^c. Number of factors?
A1. (a+1)(b+1)(c+1).

Q2. How many factors does 72 have?
A2. 72 = 2³·3² → 4·3 = 12 factors.

Q3. Product of all factors of N?
A3. N^(d/2), where d = number of factors.

Q4. Sum of factors of 12?
A4. (2³−1)/1 × (3²−1)/2 = 7 × 4 = 28.

Q5. Which numbers have an odd number of factors?
A5. Perfect squares (the factor √N is unpaired).

Q6. Divisibility rule for 8?
A6. The last three digits form a number divisible by 8.

Q7. Divisibility rule for 11?
A7. (Sum of odd-position digits) − (sum of even-position digits) is divisible by 11.

Q8. Divisibility rule for 9?
A8. The digit sum is divisible by 9.

Q9. To test divisibility by 12, which factors do you check?
A9. 3 and 4 (co-prime) — not 2 and 6.

Q10. Number of factors is odd — what does that mean?
A10. N is a perfect square.

Q11. The classic factor-count mistake?
A11. Forgetting the "+1": the count is (a+1)(b+1)..., not a·b·c.

Q12. Divisibility rule for 4?
A12. The last two digits form a number divisible by 4.

Number System is CAT's most conceptual QA area — few formulas, lots of insight — and factors/divisibility is its foundation. One idea unlocks most of it: prime factorisation. Once you can write any number as a product of prime powers, you can instantly count its factors, sum them, find how many end in zero, and test divisibility — all without heavy computation. This is high-frequency, high-return CAT territory.

What this tests

Structural understanding of integers — prime factorisation and its consequences for counting factors, divisibility rules, and number properties.

The method

Beginner — prime factorisation is the master key

Write N as p₁^a · p₂^b · p₃^c … (primes to powers). From this single form:

• Number of factors = (a+1)(b+1)(c+1)… — add one to each exponent and multiply.
• Sum of factors = [(p₁^(a+1) − 1)/(p₁ − 1)] × … for each prime.
• Product of all factors = N^(number of factors / 2).
  Example: 72 = 2³ · 3². Factors = (3+1)(2+1) = 12.

Intermediate — divisibility rules and their logic

Know the quick tests: divisible by 2 (even), 3 (digit sum ÷ 3), 4 (last two digits ÷ 4), 5 (ends 0/5), 6 (by 2 and 3), 8 (last three digits ÷ 8), 9 (digit sum ÷ 9), 11 (alternating digit sum ÷ 11). For divisibility by a composite like 12, check its coprime factors (4 and 3) — a number divisible by both 4 and 3 is divisible by 12. Never test 12 by "2 and 6" (not coprime — 2 and 6 share a factor, so it fails).

Advanced — the FAST CAT approach: manipulate exponents, not numbers

• Perfect squares have an odd number of factors (every exponent even → (even+1) is odd), so counts of squares/non-squares fall out from the factor formula.
• Number of even factors of N = total factors of N minus factors of the odd part; odd factors = factors of N with all 2s removed.
• For "how many trailing zeros in a product/factorial", count the powers of 5 (they're scarcer than 2s): zeros in n! = ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + …
• To find "factors that are multiples of k", divide out k and count factors of the quotient.

Worked example

Question: How many factors of 3600 are perfect squares?
3600 = 36 × 100 = (2² · 3²)(2² · 5²) = 2⁴ · 3² · 5². A factor is a perfect square iff each prime's exponent is even. For 2⁴: even exponents available are 0, 2, 4 → 3 choices. For 3²: 0, 2 → 2 choices. For 5²: 0, 2 → 2 choices. Perfect-square factors = 3 × 2 × 2 = 12. No brute listing needed — the even-exponent condition, applied to the factorisation, counts them directly. Answer: 12.

CAT relevance

Number System appears in 3–5 QA questions, and factor/divisibility ideas recur inside HCF-LCM, remainders, and even combinatorics. The factor-counting and trailing-zero techniques are frequent, and the "manipulate exponents" mindset is the signature of Number System mastery that distinguishes strong QA scorers.

Speed tricks & shortcuts

• Prime-factorise first — it answers factor count, sum, and divisibility all at once.
• Factors = product of (exponent + 1).
• Perfect squares ↔ all exponents even; perfect cubes ↔ all divisible by 3.
• Trailing zeros = power of 5 (in factorials, sum the floor-divisions by 5, 25, 125…).
• Mnemonic — "Factorise, then the exponents do the work."

Key formulas

• Natural, whole, integer, rational (p/q), irrational, real, prime, composite numbers.
• Sum of first n naturals = n(n+1)/2; of squares = n(n+1)(2n+1)/6; of cubes = [n(n+1)/2]².
• Even + even = even; odd + odd = even; product of two odds is odd.
• A prime > 3 is of the form 6k ± 1.
• Number of primes is infinite; 2 is the only even prime.
• (aⁿ − bⁿ) is divisible by (a − b) for all n; by (a + b) for even n.

Usage: use the standard series sums and divisibility identities to shortcut number problems.

Worked Example

Problem: For N = 360: (a) find the total number of factors, (b) the number of EVEN factors, and (c) the sum of all factors.

Solution:
First find the prime factorisation, the foundation of all factor questions.
360 = 8 × 45 = 2^3 × 9 × 5 = 2^3 × 3^2 × 5^1.

(a) Total number of factors = product of (each exponent + 1).
= (3 + 1)(2 + 1)(1 + 1) = 4 × 3 × 2 = 24 factors.
(Reason: a factor picks 0–3 twos, 0–2 threes, 0–1 fives → 4 × 3 × 2 choices.)

(b) Even factors = factors that include at least one 2. Count = total factors − odd factors. Odd factors come from 3^2 × 5^1 only (no 2): (2 + 1)(1 + 1) = 3 × 2 = 6. So even factors = 24 − 6 = 18.
(Alternatively: an even factor uses 2^1, 2^2, or 2^3 — 3 choices for the power of 2 — times 3 × 2 for the rest = 3 × 3 × 2 = 18. Same answer.)

(c) Sum of all factors uses the divisor-sum formula: for p^a × q^b × r^c, the sum = (p^0+...+p^a)(q^0+...+q^b)(r^0+...+r^c).
= (1 + 2 + 4 + 8)(1 + 3 + 9)(1 + 5)
= (15)(13)(6)
= 15 × 78 = 1170.

Answer: 360 = 2^3 × 3^2 × 5. (a) 24 factors, (b) 18 even factors (24 − 6 odd), (c) sum of factors = 15 × 13 × 6 = 1170.