Base | Representation |
---|---|
bin | 11011010010101101101010… |
… | …101001101000101000100111 |
3 | 120202000001202220022021110211 |
4 | 123102231222221220220213 |
5 | 111213111034010224201 |
6 | 1103142302004200251 |
7 | 34166054410224535 |
oct | 3322555251505047 |
9 | 522001686267424 |
10 | 120033240320551 |
11 | 352788a19a0008 |
12 | 115672b5399087 |
13 | 51c9116c09cbc |
14 | 218d8d599b155 |
15 | dd2517994651 |
hex | 6d2b6aa68a27 |
120033240320551 has 4 divisors (see below), whose sum is σ = 120033304181992. Its totient is φ = 120033176459112.
The previous prime is 120033240320489. The next prime is 120033240320573. The reversal of 120033240320551 is 155023042330021.
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is not a de Polignac number, because 120033240320551 - 211 = 120033240318503 is a prime.
It is a Duffinian number.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (120033240320051) 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, 29023080 + ... + 32899933.
It is an arithmetic number, because the mean of its divisors is an integer number (30008326045498).
Almost surely, 2120033240320551 is an apocalyptic number.
120033240320551 is a deficient number, since it is larger than the sum of its proper divisors (63861441).
120033240320551 is an equidigital number, since it uses as much as digits as its factorization.
120033240320551 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 63861440.
The product of its (nonzero) digits is 21600, while the sum is 31.
Adding to 120033240320551 its reverse (155023042330021), we get a palindrome (275056282650572).
The spelling of 120033240320551 in words is "one hundred twenty trillion, thirty-three billion, two hundred forty million, three hundred twenty thousand, five hundred fifty-one".
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