Base | Representation |
---|---|
bin | 111111111100011110… |
… | …1000011010100010001 |
3 | 111010110010012101010112 |
4 | 1333320331003110101 |
5 | 4222213102140423 |
6 | 143030112011105 |
7 | 12630636013664 |
oct | 1777075032421 |
9 | 433403171115 |
10 | 137320740113 |
11 | 53267a83577 |
12 | 22744530a95 |
13 | cc4578a80b |
14 | 69098776db |
15 | 388a8d6a78 |
hex | 1ff8f43511 |
137320740113 has 2 divisors, whose sum is σ = 137320740114. Its totient is φ = 137320740112.
The previous prime is 137320740091. The next prime is 137320740119. The reversal of 137320740113 is 311047023731.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 114235888144 + 23084851969 = 337988^2 + 151937^2 .
It is a cyclic number.
It is not a de Polignac number, because 137320740113 - 216 = 137320674577 is a prime.
It is a Sophie Germain prime.
It is a Curzon number.
It is a self number, because there is not a number n which added to its sum of digits gives 137320740113.
It is not a weakly prime, because it can be changed into another prime (137320740119) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 68660370056 + 68660370057.
It is an arithmetic number, because the mean of its divisors is an integer number (68660370057).
Almost surely, 2137320740113 is an apocalyptic number.
It is an amenable number.
137320740113 is a deficient number, since it is larger than the sum of its proper divisors (1).
137320740113 is an equidigital number, since it uses as much as digits as its factorization.
137320740113 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 10584, while the sum is 32.
Adding to 137320740113 its reverse (311047023731), we get a palindrome (448367763844).
The spelling of 137320740113 in words is "one hundred thirty-seven billion, three hundred twenty million, seven hundred forty thousand, one hundred thirteen".
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