Base | Representation |
---|---|
bin | 101100111111111010… |
… | …011011010111111011 |
3 | 11121201021210220202011 |
4 | 230333322123113323 |
5 | 1242423112440324 |
6 | 34110235150351 |
7 | 3330224024611 |
oct | 547772332773 |
9 | 147637726664 |
10 | 48316921339 |
11 | 19544708155 |
12 | 94452a53b7 |
13 | 47301694ab |
14 | 24a4db16b1 |
15 | 13cbc34c94 |
hex | b3fe9b5fb |
48316921339 has 2 divisors, whose sum is σ = 48316921340. Its totient is φ = 48316921338.
The previous prime is 48316921291. The next prime is 48316921361. The reversal of 48316921339 is 93312961384.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 48316921339 - 213 = 48316913147 is a prime.
It is a super-2 number, since 2×483169213392 (a number of 22 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 48316921292 and 48316921301.
It is not a weakly prime, because it can be changed into another prime (48316921379) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 24158460669 + 24158460670.
It is an arithmetic number, because the mean of its divisors is an integer number (24158460670).
Almost surely, 248316921339 is an apocalyptic number.
48316921339 is a deficient number, since it is larger than the sum of its proper divisors (1).
48316921339 is an equidigital number, since it uses as much as digits as its factorization.
48316921339 is an evil number, because the sum of its binary digits is even.
The product of its digits is 839808, while the sum is 49.
The spelling of 48316921339 in words is "forty-eight billion, three hundred sixteen million, nine hundred twenty-one thousand, three hundred thirty-nine".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.131 sec. • engine limits •