Base | Representation |
---|---|
bin | 10011001000110010001… |
… | …010010010101111001001 |
3 | 11122201111200120202110122 |
4 | 103020302022102233021 |
5 | 133021311444122221 |
6 | 2444052130153025 |
7 | 164004304514450 |
oct | 23106212225711 |
9 | 4581450522418 |
10 | 1315101551561 |
11 | 4678055a5416 |
12 | 192a60a22775 |
13 | 970239ab275 |
14 | 47919085b97 |
15 | 2431eae8eab |
hex | 13232292bc9 |
1315101551561 has 4 divisors (see below), whose sum is σ = 1502973201792. Its totient is φ = 1127229901332.
The previous prime is 1315101551443. The next prime is 1315101551593. The reversal of 1315101551561 is 1651551015131.
It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-1315101551561 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (1315101571561) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 93935825105 + ... + 93935825118.
It is an arithmetic number, because the mean of its divisors is an integer number (375743300448).
Almost surely, 21315101551561 is an apocalyptic number.
It is an amenable number.
1315101551561 is a deficient number, since it is larger than the sum of its proper divisors (187871650231).
1315101551561 is an equidigital number, since it uses as much as digits as its factorization.
1315101551561 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 187871650230.
The product of its (nonzero) digits is 11250, while the sum is 35.
Adding to 1315101551561 its reverse (1651551015131), we get a palindrome (2966652566692).
The spelling of 1315101551561 in words is "one trillion, three hundred fifteen billion, one hundred one million, five hundred fifty-one thousand, five hundred sixty-one".
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