Why this topic matters
Almost every placement, bank and SSC paper has two or three number-system questions. They look hard but follow a handful of rules. Learn the rules once and most of these questions take under a minute.
Divisibility rules
Divisor | Rule | Example |
|---|---|---|
2 | last digit is even | 3,458 |
3 | sum of digits divisible by 3 | 4,521 (4 + 5 + 2 + 1 = 12) |
4 | last two digits divisible by 4 | 7,316 (16) |
5 | last digit 0 or 5 | 2,045 |
8 | last three digits divisible by 8 | 21,104 (104) |
9 | sum of digits divisible by 9 | 8,271 (18) |
11 | difference between sums of alternate digits is 0 or a multiple of 11 | 918,082: (9 + 8 + 8) − (1 + 0 + 2) = 22 |
For 6, 12, 15 and other composite divisors, check co-prime factors: a number is divisible by 12 when it is divisible by both 3 and 4.
Remainders
Remainders can be multiplied and added. The remainder of a product equals the remainder of the product of the remainders.
Example. Remainder when 15 × 17 × 19 is divided by 7. The remainders are 1, 3 and 5. 1 × 3 × 5 = 15, and 15 leaves 1 when divided by 7.
If N leaves remainder r when divided by d, then N = dk + r. So 2N = 2dk + 2r, and the remainder of 2N is the remainder of 2r.
Unit digits and cycles
The unit digit of a power depends only on the unit digit of the base, and it repeats in a cycle.
Last digit | Cycle of unit digits | Cycle length |
|---|---|---|
2 | 2, 4, 8, 6 | 4 |
3 | 3, 9, 7, 1 | 4 |
7 | 7, 9, 3, 1 | 4 |
8 | 8, 4, 2, 6 | 4 |
4 | 4, 6 | 2 |
9 | 9, 1 | 2 |
0, 1, 5, 6 | the digit itself | 1 |
Method: divide the power by the cycle length. A remainder of r means the r-th term of the cycle; a remainder of 0 means the last term.
Example. Unit digit of 7⁹⁵. 95 = 4 × 23 + 3, so take the 3rd term of 7, 9, 3, 1: 3.
The same idea gives remainders. 2ⁿ mod 5 cycles through 2, 4, 3, 1, so 2³¹ mod 5 is the 3rd term: 3.
Counting factors
Write the number as a product of prime powers, N = pᵃ × qᵇ × rᶜ. Then the number of factors is (a + 1)(b + 1)(c + 1).
Example. 360 = 2³ × 3² × 5¹ has (3 + 1)(2 + 1)(1 + 1) = 24 factors.
Trailing zeros of n!
Each trailing zero needs a 10 = 2 × 5, and fives are scarcer than twos. Count them: ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + …
Example. 100! has 20 + 4 = 24 trailing zeros.
Useful sums
1 + 2 + … + n = n(n + 1)/2
1² + 2² + … + n² = n(n + 1)(2n + 1)/6
Sum of the first n odd numbers = n²
Shortcuts and tricks
Every question here can be done by direct calculation. In a timed test, these shortcuts reach the same answer without handling large numbers.
Remainder by 9 from the digit sum
Example: the remainder when 7,384,152 is divided by 9.
Standard way: divide. 9 × 820,461 = 7,384,149, so the remainder is 3.
Shortcut: add the digits: 7 + 3 + 8 + 4 + 1 + 5 + 2 = 30, then 3 + 0 = 3. The remainder is 3.
Reduce before you multiply
Example: the remainder when 23 × 34 × 45 is divided by 11.
Standard way: 23 × 34 × 45 = 35,190, and 35,190 = 11 × 3,199 + 1, so the remainder is 1.
Shortcut: replace each number by its remainder first: 23 → 1, 34 → 1, 45 → 1. The product 1 × 1 × 1 leaves 1.
Negative remainders
Example: the remainder when 2¹⁰⁰ is divided by 3.
Standard way: powers of 2 leave remainders 2, 1, 2, 1, … on division by 3, so every even power leaves 1.
Shortcut: 2 is one less than 3, so it behaves like −1. 2¹⁰⁰ behaves like (−1)¹⁰⁰ = 1.
Unit digit of a product
Example: the unit digit of 1347 × 2458 × 963.
Standard way: multiply out the whole product and read the last digit.
Shortcut: only last digits matter: 7 × 8 = 56 → 6, then 6 × 3 = 18 → 8.
Counting factors
Example: how many factors does 72 have?
Standard way: list the factor pairs: 1 × 72, 2 × 36, 3 × 24, 4 × 18, 6 × 12, 8 × 9. That is 12 factors.
Shortcut: 72 = 2³ × 3², so the count is (3 + 1)(2 + 1) = 12.
Exam tips
To find the largest k-digit number divisible by d, divide the largest k-digit number by d and subtract the remainder.
To find the number to add so that N becomes divisible by d, subtract N's remainder from d.
Check answer options against a divisibility rule before calculating anything.