Number System & Divisibility

Elementary Mathematics (100 questions, 100 marks, 2 hours) is a scoring paper for IMA/INA/AFA. Number system and divisibility are the foundation — fast tests here save time across the whole paper.

Number families

• Natural (1,2,3...), whole (0,1,2...), integers (...,-1,0,1,...), rational (p/q form), irrational (√2, π), real (all of these).
• A prime has exactly two factors; 1 is neither prime nor composite; 2 is the only even prime.

Divisibility rules

Divisor	Test
2	last digit even
3	digit sum divisible by 3
4	last two digits form a multiple of 4
5	ends in 0 or 5
8	last three digits form a multiple of 8
9	digit sum divisible by 9
11	(sum at odd places - sum at even places) divisible by 11

Factors and standard sums

• If N = p^a x q^b x r^c, the number of factors = (a+1)(b+1)(c+1).
• Sum of first n naturals = n(n+1)/2; sum of squares = n(n+1)(2n+1)/6; sum of cubes = [n(n+1)/2]^2.
• Unit digits repeat in cycles of 4 (powers of 2 give 2, 4, 8, 6, ...).

Exam Tricks & Tips

• 🎯 For factor-count questions, prime-factorise, then apply (a+1)(b+1)(c+1).
• 🎯 Divisible by 6 = divisible by BOTH 2 and 3; by 12 = by both 3 and 4.
• 🎯 Find the unit digit of a large power by reducing the exponent mod 4 (its cyclicity).
• 🎯 The digit-sum trick (casting out nines) instantly checks divisibility by 3 and 9.
• 🎯 Remember 2 is the only even prime — questions often exploit this exception.
• ❌ Common mistake: treating 1 as a prime number — 1 is neither prime nor composite.

Expected exam pattern

Divisibility, factor-count, unit-digit and remainder questions worth 6-10 marks; quick marks once the rules are automatic.

Quick recap

Know the number families, the divisibility tests, the (a+1)(b+1)(c+1) factor formula, the standard sums, and unit-digit cyclicity.

Cover the answer, recall, then check. 12 cards on number system.

Q1. Divisibility test for 11?
A1. (Sum of digits at odd places - sum at even places) is 0 or a multiple of 11.

Q2. Number of factors of N = p^a x q^b x r^c?
A2. (a+1)(b+1)(c+1).

Q3. How many factors does 72 have?
A3. 72 = 2^3 x 3^2, so (3+1)(2+1) = 12 factors.

Q4. Sum of the first n natural numbers?
A4. n(n+1)/2.

Q5. Sum of squares of the first n naturals?
A5. n(n+1)(2n+1)/6.

Q6. Is 1 prime, composite, or neither?
A6. Neither prime nor composite.

Q7. Which is the only even prime number?
A7. 2.

Q8. Unit digit of 2^43?
A8. Cycle is 2,4,8,6 (length 4); 43 mod 4 = 3, so the unit digit is 8.

Q9. Divisibility test for 8?
A9. The last three digits form a number divisible by 8.

Q10. When is a number divisible by 6?
A10. When it is divisible by both 2 and 3.

Q11. Sum of cubes of the first n naturals?
A11. [n(n+1)/2]^2.

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

The Number System is the foundation of CDS Elementary Mathematics — every arithmetic topic builds on classifying numbers and testing divisibility. It rewards a small set of rules that let you simplify problems without brute-force calculation.

Core idea / what this tests

This tests how numbers are classified (natural, integer, rational, prime, etc.) and the divisibility rules, remainders, and unit-digit patterns used to answer questions quickly. The goal is to replace long division with instant tests.

Deep explanation

Beginner — classifying numbers

• Natural numbers (1, 2, 3…), whole numbers (0, 1, 2…), integers (…−2, −1, 0, 1, 2…).
• Rational numbers can be written as p/q (q ≠ 0); irrational numbers (√2, π) cannot.
• Prime numbers have exactly two factors (1 and itself); 1 is neither prime nor composite; 2 is the only even prime.

Intermediate — divisibility rules

Memorise these tests:

• 2: last digit even. 5: last digit 0 or 5. 10: ends in 0.
• 3: digit sum divisible by 3. 9: digit sum divisible by 9.
• 4: last two digits divisible by 4. 8: last three digits divisible by 8.
• 6: divisible by both 2 and 3. 11: (sum of odd-position digits − sum of even-position digits) is 0 or a multiple of 11.

Advanced — remainders and unit digits

• Remainder theorem for products: the remainder of a product equals the product of the remainders (then take mod again). E.g., remainder of (17 × 23) ÷ 5 = (2 × 3) mod 5 = 6 mod 5 = 1.
• Unit-digit cyclicity: powers repeat their unit digit in cycles of 4. For 7: 7¹=7, 7²=9, 7³=3, 7⁴=1, then repeats. To find the unit digit of 7⁵³, divide the power 53 by 4 (remainder 1) → same unit digit as 7¹ = 7.
• Number of factors: if N = aᵖ × bᵠ (prime factorisation), the number of factors = (p+1)(q+1).

Worked example

Find the unit digit of 3⁶⁴.

Step 1 — find the cycle of 3: 3¹=3, 3²=9, 3³=27→7, 3⁴=81→1, then repeats every 4.
Step 2 — divide the exponent 64 by the cycle length 4: 64 ÷ 4 = 16 remainder 0.
Step 3 — a remainder of 0 corresponds to the last number in the cycle (the 4th power), whose unit digit is 1.
Answer: the unit digit of 3⁶⁴ is 1. No large computation needed — cyclicity does the work.

CDS relevance

Number-system questions are frequent in CDS and are the fastest to solve once the rules are internalised. Divisibility tests and unit-digit tricks save precious minutes under the 2-hour, 100-question maths paper, and they recur in HCF/LCM and simplification problems.

Exam tricks & shortcuts

• Divisibility by 3/9: just add the digits — never divide.
• Unit digit of powers: cycle length is 4 (except digits 0,1,5,6 which stay fixed); use exponent mod 4, with remainder 0 → the 4th value.
• Factor count: prime-factorise, add 1 to each exponent, multiply.
• Mnemonic for divisibility by 11: "odd minus even = 0 or 11's multiple."

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: Find the least number that must be added to 1056 to make it exactly divisible by 23.

Solution:
When a number is not divisible by a divisor, dividing leaves a remainder. To reach the next multiple of the divisor, we add (divisor − remainder).

Step 1 — Divide 1056 by 23:
23 × 45 = 1035, and 23 × 46 = 1058.
So 1056 lies between 1035 and 1058. Dividing: 1056 = 23 × 45 + 21, giving a remainder of 21.

Step 2 — The next multiple of 23 after 1056 is 1058. The number to add is:
1058 − 1056 = 2.
(Equivalently, divisor − remainder = 23 − 21 = 2.)

Step 3 — Check: 1056 + 2 = 1058 = 23 × 46 ✓, which is exactly divisible by 23.

Answer: 2 must be added.