The number 59321
59321 is a composite number with 4 divisors. Its prime factorisation is 137 × 433. Divisors (4): 1, 137, 433, 59321.
Pick any whole number from 1 to 100,000.
How to work it out
- 1
What is special about 59321: a semiprime: 137 × 433, the product of two primes.
- 2
59321 is composite. Its prime factorisation is
- 3
The divisors add up to 59892; without 59321 itself that is 571, so 59321 is deficient (571 is less than 59321). Euler's totient φ(59321) = 58752 numbers up to 59321 share no factor with it.
- 4
The digits of 59321 add up to 20, and repeating that until one digit is left gives the digital root 2. The Collatz sequence (halve if even, else 3n+1) takes 96 steps to reach 1.
LaTeX source of the formulas on this page
\[ 59321 = 137 \times 433 \]
59321 at a glance
| Property | Value |
|---|---|
| Type | composite |
| Number of divisors | 4 |
| Sum of divisors | 59892 |
| Euler totient φ(n) | 58752 |
| Square | 3518981041 |
| Cube | 208749474333161 |
| Square root | 243.559028 |
| Digit sum / digital root | 20 / 2 |
| Collatz steps to 1 | 96 |
| Scientific notation | 5.9321 × |
59321 in other forms
| Form | Value |
|---|---|
| In words | fifty-nine thousand three hundred and twenty-one |
| Binary (base 2) | 1110011110111001 |
| Octal (base 8) | 163671 |
| Hexadecimal (base 16) | E7B9 |
| Base 3 | 10000101002 |
| As time | 59321 seconds = 16 h 28 min 41 s |
Is 59321 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 | No (remainder 2) | A number is divisible by 3 if the digits add up to a multiple of 3. |
| 4 | No (remainder 1) | A number is divisible by 4 if the last two digits make a multiple of 4. |
| 5 | No (remainder 1) | A number is divisible by 5 if the last digit is 0 or 5. |
| 6 | No (remainder 5) | A number is divisible by 6 if it is divisible by both 2 and 3. |
| 7 | No (remainder 3) | 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 1) | A number is divisible by 8 if the last three digits make a multiple of 8. |
| 9 | No (remainder 2) | A number is divisible by 9 if the digits add up to a multiple of 9. |
| 10 | No (remainder 1) | A number is divisible by 10 if the last digit is 0. |
| 11 | No (remainder 9) | A number is divisible by 11 if the alternating sum of the digits is a multiple of 11. |
| 12 | No (remainder 5) | A number is divisible by 12 if it is divisible by both 3 and 4. |
Questions
Is 59321 a prime number?
No. 59321 = 137 × 433, so it is divisible by 137.
What are the factors of 59321?
1, 137, 433, 59321.
What is 59321 in binary?
1110011110111001 (hexadecimal E7B9, octal 163671).
What is the square root of 59321?
About 243.5590.
Is 59321 divisible by 3?
No. Its digits add up to 20, which is not a multiple of 3.
What is special about the number 59321?
59321 is a semiprime: 137 × 433, the product of two primes.