60-second revision
- 1 is neither prime nor composite; 2 is the only even prime; there are 25 primes below 100.
- Divisible by 11: (sum of alternate digits) minus (sum of the others) is 0 or a multiple of 11.
- Unit digits of powers repeat every 4: divide the power by 4 and use the remainder (remainder 0 means the 4th term).
- If N = 2^a × 3^b × 5^c, the number of factors is (a + 1)(b + 1)(c + 1).
- Sum of first n odd numbers = n²; sum of first n natural numbers = n(n + 1)/2.
- Trailing zeros in n! = n/5 + n/25 + n/125 + ... (whole parts only).
- If N leaves remainder r when divided by d, and k divides d, then N leaves r mod k when divided by k.
- To test whether n is prime, check divisibility only by primes up to the square root of n.
Types of numbers
| Type | Meaning | Examples |
|---|---|---|
| Natural | Counting numbers | 1, 2, 3, ... |
| Whole | Natural numbers plus 0 | 0, 1, 2, ... |
| Integers | Whole numbers and their negatives | ..., -2, -1, 0, 1, 2, ... |
| Rational | Can be written as p/q with q not 0; decimals terminate or repeat | 3/4, -5, 0.25, 0.333... |
| Irrational | Decimals never end and never repeat | sqrt 2, sqrt 5, pi |
| Prime | Exactly two factors, 1 and itself | 2, 3, 5, 7, 11, 13 |
| Composite | More than two factors | 4, 6, 9, 15 |
| Co-prime | Two numbers whose HCF is 1 (need not be prime) | 8 and 15 |
1 is neither prime nor composite. 2 is the only even prime. There are 25 primes below 100 and 15 below 50. To check whether a number like 391 is prime, test only the primes up to its square root (here up to 19): 391 = 17 × 23, so it is composite.
Divisibility rules
| Divisor | Rule | Example |
|---|---|---|
| 2 | Last digit even | 3,578 |
| 3 | Sum of digits divisible by 3 | 4,113 (sum 9) |
| 4 | Last two digits divisible by 4 | 7,316 (16) |
| 5 | Ends in 0 or 5 | 9,435 |
| 6 | Divisible by both 2 and 3 | 8,412 |
| 8 | Last three digits divisible by 8 | 51,264 (264) |
| 9 | Sum of digits divisible by 9 | 7,452 (sum 18) |
| 11 | Alternate-digit sums differ by 0 or a multiple of 11 | 3,72,416 |
| 12 | Divisible by both 3 and 4 | 5,124 |
Worked example (11): for 3,72,416 take digits from the left with alternate signs: 3 - 7 + 2 - 4 + 1 - 6 = -11, a multiple of 11, so the number is divisible by 11 (3,72,416 = 11 × 33,856).
Missing-digit questions: for 9 or 3, add the known digits and choose the digit that makes the total a multiple. For 8, only the last three digits matter.
Unit digit of a power
| Last digit of base | Cycle of unit digits | Cycle length |
|---|---|---|
| 0, 1, 5, 6 | Always the same digit | 1 |
| 4 | 4, 6 | 2 |
| 9 | 9, 1 | 2 |
| 2 | 2, 4, 8, 6 | 4 |
| 3 | 3, 9, 7, 1 | 4 |
| 7 | 7, 9, 3, 1 | 4 |
| 8 | 8, 4, 2, 6 | 4 |
Method: divide the power by 4. Remainder 1, 2, 3 picks the 1st, 2nd, 3rd digit of the cycle; remainder 0 picks the 4th. Example: unit digit of 3^58. 58 = 4 × 14 + 2, so take the 2nd digit of 3, 9, 7, 1, which is 9.
Factors of a number
Write the number as a product of prime powers. If N = p^a × q^b × r^c, the number of factors is (a + 1)(b + 1)(c + 1). Example: 720 = 2^4 × 3² × 5, so it has 5 × 3 × 2 = 30 factors (1 and 720 included).
Sum formulas
| Series | Sum | Example |
|---|---|---|
| 1 + 2 + ... + n | n(n + 1)/2 | 1 to 50 = 1,275 |
| 1² + 2² + ... + n² | n(n + 1)(2n + 1)/6 | 1² to 10² = 385 |
| 1³ + 2³ + ... + n³ | [n(n + 1)/2]² | 1³ to 5³ = 225 |
| First n odd numbers | n² | 1 + 3 + ... + 19 = 100 |
| First n even numbers | n(n + 1) | 2 + 4 + ... + 20 = 110 |
Remainders
- Dividend = divisor × quotient + remainder. The remainder is always smaller than the divisor.
- Divisor-of-divisor rule: a number leaves 47 when divided by 72. Since 9 divides 72, the remainder on dividing by 9 is 47 mod 9 = 2.
- Power remainders: find a power that leaves remainder 1. 2³ = 8 leaves 1 on division by 7, so 2^50 = (2³)^16 × 2², giving remainder 4.
Trailing zeros in a factorial
Each trailing zero needs a pair 2 × 5, and 5s are always the scarcer factor. Count the 5s: zeros in n! = [n/5] + [n/25] + [n/125] + ..., keeping only whole parts. Example: 250! has 50 + 10 + 2 = 62 trailing zeros.
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