The number 49343
49343 is a composite number with 12 divisors. Its prime factorisation is × 19 × 53. Divisors (12): 1, 7, 19, 49, 53, 133, 371, 931, 1007, 2597, 7049, 49343.
Pick any whole number from 1 to 100,000.
How to work it out
- 1
49343 has no famous special property in the usual lists (prime, square, cube, triangular, Fibonacci, palindrome, perfect), but it is still a composite number.
- 2
49343 is composite. Its prime factorisation is
- 3
The divisors add up to 61560; without 49343 itself that is 12217, so 49343 is deficient (12217 is less than 49343). Euler's totient φ(49343) = 39312 numbers up to 49343 share no factor with it.
- 4
The digits of 49343 add up to 23, and repeating that until one digit is left gives the digital root 5. The Collatz sequence (halve if even, else 3n+1) takes 158 steps to reach 1.
LaTeX source of the formulas on this page
\[ 49343 = 7^{2} \times 19 \times 53 \]49343 at a glance
| Property | Value |
|---|---|
| Type | composite |
| Number of divisors | 12 |
| Sum of divisors | 61560 |
| Euler totient φ(n) | 39312 |
| Square | 2434731649 |
| Cube | 120136963756607 |
| Square root | 222.132843 (= 7√1007) |
| Digit sum / digital root | 23 / 5 |
| Collatz steps to 1 | 158 |
| Scientific notation | 4.9343 × |
49343 in other forms
| Form | Value |
|---|---|
| In words | forty-nine thousand three hundred and forty-three |
| Binary (base 2) | 1100000010111111 |
| Octal (base 8) | 140277 |
| Hexadecimal (base 16) | C0BF |
| Base 3 | 2111200112 |
| As time | 49343 seconds = 13 h 42 min 23 s |
Is 49343 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 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 5) | A number is divisible by 6 if it is divisible by both 2 and 3. |
| 7 | Yes (49343 ÷ 7 = 7049) | 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 7) | 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 11) | A number is divisible by 12 if it is divisible by both 3 and 4. |
Questions
Is 49343 a prime number?
No. 49343 = × 19 × 53, so it is divisible by 7.
What are the factors of 49343?
1, 7, 19, 49, 53, 133, 371, 931, 1007, 2597, 7049, 49343.
What is 49343 in binary?
1100000010111111 (hexadecimal C0BF, octal 140277).
What is the square root of 49343?
About 222.1328, exactly 7√1007.
Is 49343 divisible by 3?
No. Its digits add up to 23, which is not a multiple of 3.
What is special about the number 49343?
It is a composite number with 12 divisors.