Base | Representation |
---|---|
bin | 10000110000011101110… |
… | …111001011110111101011 |
3 | 11002002100122200221212212 |
4 | 100300131313023313223 |
5 | 122331334333204011 |
6 | 2241003224555335 |
7 | 146124352355435 |
oct | 20603567136753 |
9 | 4062318627785 |
10 | 1151552241131 |
11 | 404409245856 |
12 | 167218622b4b |
13 | 8478b2cb776 |
14 | 3da42019655 |
15 | 1ee4b75e08b |
hex | 10c1ddcbdeb |
1151552241131 has 4 divisors (see below), whose sum is σ = 1152629466000. Its totient is φ = 1150475016264.
The previous prime is 1151552241127. The next prime is 1151552241157. The reversal of 1151552241131 is 1311422551511.
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 1151552241131 - 22 = 1151552241127 is a prime.
It is a super-2 number, since 2×11515522411312 (a number of 25 digits) contains 22 as substring.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (1151552241331) 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, 538610831 + ... + 538612968.
It is an arithmetic number, because the mean of its divisors is an integer number (288157366500).
Almost surely, 21151552241131 is an apocalyptic number.
1151552241131 is a deficient number, since it is larger than the sum of its proper divisors (1077224869).
1151552241131 is a wasteful number, since it uses less digits than its factorization.
1151552241131 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 1077224868.
The product of its digits is 6000, while the sum is 32.
Adding to 1151552241131 its reverse (1311422551511), we get a palindrome (2462974792642).
The spelling of 1151552241131 in words is "one trillion, one hundred fifty-one billion, five hundred fifty-two million, two hundred forty-one thousand, one hundred thirty-one".
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