Base | Representation |
---|---|
bin | 1110000010110110110111… |
… | …1001001010110001100010 |
3 | 2000200021011000212122022012 |
4 | 3200231231321022301202 |
5 | 4011001212242211002 |
6 | 52502023334112522 |
7 | 3152444243611352 |
oct | 340555571126142 |
9 | 60607130778265 |
10 | 15442251132002 |
11 | 4a14027204792 |
12 | 189498a7ab742 |
13 | 880276105436 |
14 | 3b55a21d2b62 |
15 | 1bba4d4b1d52 |
hex | e0b6de4ac62 |
15442251132002 has 4 divisors (see below), whose sum is σ = 23163376698006. Its totient is φ = 7721125566000.
The previous prime is 15442251131977. The next prime is 15442251132017. The reversal of 15442251132002 is 20023115224451.
15442251132002 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a semiprime because it is the product of two primes.
It can be written as a sum of positive squares in only one way, i.e., 14592331240081 + 849919891921 = 3819991^2 + 921911^2 .
It is a self number, because there is not a number n which added to its sum of digits gives 15442251132002.
It is an unprimeable number.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 3860562782999 + ... + 3860562783002.
Almost surely, 215442251132002 is an apocalyptic number.
15442251132002 is a deficient number, since it is larger than the sum of its proper divisors (7721125566004).
15442251132002 is an equidigital number, since it uses as much as digits as its factorization.
15442251132002 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 7721125566003.
The product of its (nonzero) digits is 19200, while the sum is 32.
Adding to 15442251132002 its reverse (20023115224451), we get a palindrome (35465366356453).
The spelling of 15442251132002 in words is "fifteen trillion, four hundred forty-two billion, two hundred fifty-one million, one hundred thirty-two thousand, two".
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