Base | Representation |
---|---|
bin | 1000110100111000001… |
… | …1010100110111110011 |
3 | 112111101120220022000212 |
4 | 2031032003110313303 |
5 | 4441021114201011 |
6 | 153354232534335 |
7 | 13645421131664 |
oct | 2151603246763 |
9 | 474346808025 |
10 | 151633350131 |
11 | 5934208929a |
12 | 254798513ab |
13 | 113b6a91023 |
14 | 74a668736b |
15 | 3e271ba08b |
hex | 234e0d4df3 |
151633350131 has 4 divisors (see below), whose sum is σ = 151634285664. Its totient is φ = 151632414600.
The previous prime is 151633350109. The next prime is 151633350157. The reversal of 151633350131 is 131053336151.
It is a semiprime because it is the product of two primes, and also a brilliant number, because the two primes have the same length.
It is a cyclic number.
It is not a de Polignac number, because 151633350131 - 230 = 150559608307 is a prime.
It is a super-2 number, since 2×1516333501312 (a number of 23 digits) contains 22 as substring.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (151633350101) 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, 154880 + ... + 572061.
It is an arithmetic number, because the mean of its divisors is an integer number (37908571416).
Almost surely, 2151633350131 is an apocalyptic number.
151633350131 is a deficient number, since it is larger than the sum of its proper divisors (935533).
151633350131 is an equidigital number, since it uses as much as digits as its factorization.
151633350131 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 935532.
The product of its (nonzero) digits is 12150, while the sum is 32.
Adding to 151633350131 its reverse (131053336151), we get a palindrome (282686686282).
The spelling of 151633350131 in words is "one hundred fifty-one billion, six hundred thirty-three million, three hundred fifty thousand, one hundred thirty-one".
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