Number System — Types, Divisibility, Unit Digit and Remainders

Almost every aptitude question uses numbers, so this lesson is the foundation. You will learn four things:

1. The types of numbers (and which one contains which)
2. Divisibility rules — check if a number divides another without long division
3. The unit digit (last digit) of huge powers like 7¹⁰⁵
4. Remainders — the most common "number system" question in placement tests

1. Types of numbers

Two more you must know:

• Prime number: exactly two factors, 1 and itself → 2, 3, 5, 7, 11, 13, … (2 is the only even prime; 1 is not prime).
• Composite number: more than two factors → 4, 6, 8, 9, …

Remember: there are 25 primes below 100. A number is prime if no prime up to its square root divides it. To test 97: √97 ≈ 9.8, so try 2, 3, 5, 7 — none divide 97, so 97 is prime.

2. Divisibility rules

For 12, 15, 18, 24… split into two numbers that have no common factor: 12 = 3 × 4, 15 = 3 × 5, 24 = 3 × 8. Check both rules. (Do not use 12 = 2 × 6 — 2 and 6 share the factor 2.)

Example 1

If the number 5 7 x 2 is divisible by 9, what is x?

Digit sum = 5 + 7 + x + 2 = 14 + x. The next multiple of 9 after 14 is 18, so x = 4. (27 would need x = 13, which is not a digit.)

3. Unit digit of big powers

The last digit of powers repeats in a cycle. Only the last digit of the base matters.

Method (for bases ending in 2, 3, 7, 8):

1. Divide the power by 4 and take the remainder.
2. Remainder 1, 2, 3 → the 1st, 2nd, 3rd digit of the cycle. Remainder 0 → the 4th digit.

Example 2

Find the unit digit of 7¹⁰⁵.

105 ÷ 4 leaves remainder 1 → 1st digit of 7's cycle (7, 9, 3, 1) = 7.

Example 3

Find the unit digit of 23⁴⁸ × 12⁵¹.

• 23⁴⁸ → base ends in 3, 48 ÷ 4 remainder 0 → 4th digit of (3, 9, 7, 1) = 1
• 12⁵¹ → base ends in 2, 51 ÷ 4 remainder 3 → 3rd digit of (2, 4, 8, 6) = 8
• Unit digit = 1 × 8 = 8

4. Remainders

Rule 1 — remainders multiply. The remainder of (a × b) ÷ n equals the remainder of (remainder of a × remainder of b) ÷ n. The same works for adding.

Example 4

What is the remainder when 856 × 501 is divided by 13?

• 856 ÷ 13 → 13 × 65 = 845, remainder 11
• 501 ÷ 13 → 13 × 38 = 494, remainder 7
• 11 × 7 = 77; 77 ÷ 13 → 13 × 5 = 65, remainder 12

Answer: 12. You never had to multiply the big numbers.

Rule 2 — negative remainders save time. 11 is 2 less than 13, so think of it as −2. Then (−2) × 7 = −14, and −14 + 26 = 12. Same answer, smaller numbers.

Example 5

Find the remainder when 2³¹ is divided by 5.

2⁴ = 16 leaves remainder 1 when divided by 5. 2³¹ = (2⁴)⁷ × 2³ → remainder = 1⁷ × 8 → 8 ÷ 5 → 3.

Common mistakes

• Calling 1 a prime number. It is not (it has only one factor).
• Using 12 = 2 × 6 for divisibility. The two parts must have no common factor: use 3 × 4.
• For unit digits, a remainder of 0 means the last digit of the cycle, not the first.

Quick recap

• Natural ⊂ Whole ⊂ Integers ⊂ Rational ⊂ Real. 2 is the only even prime.
• Learn divisibility rules for 2–11; split bigger numbers into co-prime parts.
• Unit digit: find the cycle, divide the power by the cycle length (4 works for every digit).
• Remainders: replace each number with its remainder first, then multiply or add.