The number 83993
83993 is a composite number with 12 divisors. Its prime factorisation is 7 × × 71. Divisors (12): 1, 7, 13, 71, 91, 169, 497, 923, 1183, 6461, 11999, 83993.
Pick any whole number from 1 to 100,000.
How to work it out
- 1
83993 has no famous special property in the usual lists (prime, square, cube, triangular, Fibonacci, palindrome, perfect), but it is still a composite number.
- 2
83993 is composite. Its prime factorisation is
- 3
The divisors add up to 105408; without 83993 itself that is 21415, so 83993 is deficient (21415 is less than 83993). Euler's totient φ(83993) = 65520 numbers up to 83993 share no factor with it.
- 4
The digits of 83993 add up to 32, and repeating that until one digit is left gives the digital root 5. The Collatz sequence (halve if even, else 3n+1) takes 89 steps to reach 1.
LaTeX source of the formulas on this page
\[ 83993 = 7 \times 13^{2} \times 71 \]83993 at a glance
| Property | Value |
|---|---|
| Type | composite |
| Number of divisors | 12 |
| Sum of divisors | 105408 |
| Euler totient φ(n) | 65520 |
| Square | 7054824049 |
| Cube | 592555836347657 |
| Square root | 289.815459 (= 13√497) |
| Digit sum / digital root | 32 / 5 |
| Collatz steps to 1 | 89 |
| Scientific notation | 8.3993 × |
83993 in other forms
| Form | Value |
|---|---|
| In words | eighty-three thousand nine hundred and ninety-three |
| Binary (base 2) | 10100100000011001 |
| Octal (base 8) | 244031 |
| Hexadecimal (base 16) | 14819 |
| Base 3 | 11021012212 |
| As time | 83993 seconds = 23 h 19 min 53 s |
Is 83993 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 3) | 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 | Yes (83993 ÷ 7 = 11999) | 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 5) | 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 8) | 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 83993 a prime number?
No. 83993 = 7 × × 71, so it is divisible by 7.
What are the factors of 83993?
1, 7, 13, 71, 91, 169, 497, 923, 1183, 6461, 11999, 83993.
What is 83993 in binary?
10100100000011001 (hexadecimal 14819, octal 244031).
What is the square root of 83993?
About 289.8155, exactly 13√497.
Is 83993 divisible by 3?
No. Its digits add up to 32, which is not a multiple of 3.
What is special about the number 83993?
It is a composite number with 12 divisors.