The number 20043
20043 is a composite number with 12 divisors. Its prime factorisation is × 17 × 131. Divisors (12): 1, 3, 9, 17, 51, 131, 153, 393, 1179, 2227, 6681, 20043.
Pick any whole number from 1 to 100,000.
How to work it out
- 1
What is special about 20043: a Harshad number: divisible by its digit sum 9.
- 2
20043 is composite. Its prime factorisation is
- 3
The divisors add up to 30888; without 20043 itself that is 10845, so 20043 is deficient (10845 is less than 20043). Euler's totient φ(20043) = 12480 numbers up to 20043 share no factor with it.
- 4
The digits of 20043 add up to 9, and repeating that until one digit is left gives the digital root 9. The Collatz sequence (halve if even, else 3n+1) takes 92 steps to reach 1.
LaTeX source of the formulas on this page
\[ 20043 = 3^{2} \times 17 \times 131 \]20043 at a glance
| Property | Value |
|---|---|
| Type | composite |
| Number of divisors | 12 |
| Sum of divisors | 30888 |
| Euler totient φ(n) | 12480 |
| Square | 401721849 |
| Cube | 8051711019507 |
| Square root | 141.573303 (= 3√2227) |
| Digit sum / digital root | 9 / 9 |
| Collatz steps to 1 | 92 |
| Scientific notation | 2.0043 × |
20043 in other forms
| Form | Value |
|---|---|
| In words | twenty thousand and forty-three |
| Binary (base 2) | 100111001001011 |
| Octal (base 8) | 47113 |
| Hexadecimal (base 16) | 4E4B |
| Base 3 | 1000111100 |
| As time | 20043 seconds = 5 h 34 min 3 s |
Is 20043 divisible by 2 to 12?
| Divisor | Answer | Rule |
|---|---|---|
| 2 | No (remainder 1) | A number is divisible by 2 if the last digit is even (0, 2, 4, 6, 8). |
| 3 | Yes (20043 ÷ 3 = 6681) | A number is divisible by 3 if the digits add up to a multiple of 3. |
| 4 | No (remainder 3) | A number is divisible by 4 if the last two digits make a multiple of 4. |
| 5 | No (remainder 3) | A number is divisible by 5 if the last digit is 0 or 5. |
| 6 | No (remainder 3) | A number is divisible by 6 if it is divisible by both 2 and 3. |
| 7 | No (remainder 2) | A number is divisible by 7 if double the last digit and subtract it from the rest: the result is a multiple of 7. |
| 8 | No (remainder 3) | A number is divisible by 8 if the last three digits make a multiple of 8. |
| 9 | Yes (20043 ÷ 9 = 2227) | A number is divisible by 9 if the digits add up to a multiple of 9. |
| 10 | No (remainder 3) | A number is divisible by 10 if the last digit is 0. |
| 11 | No (remainder 1) | A number is divisible by 11 if the alternating sum of the digits is a multiple of 11. |
| 12 | No (remainder 3) | A number is divisible by 12 if it is divisible by both 3 and 4. |
Questions
Is 20043 a prime number?
No. 20043 = × 17 × 131, so it is divisible by 3.
What are the factors of 20043?
1, 3, 9, 17, 51, 131, 153, 393, 1179, 2227, 6681, 20043.
What is 20043 in binary?
100111001001011 (hexadecimal 4E4B, octal 47113).
What is the square root of 20043?
About 141.5733, exactly 3√2227.
Is 20043 divisible by 3?
Yes. Its digits add up to 9, a multiple of 3.
What is special about the number 20043?
20043 is a Harshad number: divisible by its digit sum 9.