Base | Representation |
---|---|
bin | 11111000000110000010… |
… | …011111011100110011101 |
3 | 21112201210000112220121022 |
4 | 133000300103323212131 |
5 | 234404010230410140 |
6 | 4311004223413525 |
7 | 306652666515530 |
oct | 37006023734635 |
9 | 7481700486538 |
10 | 2131114310045 |
11 | 751889519606 |
12 | 2a5035a542a5 |
13 | 125c6a25a156 |
14 | 75209d40a17 |
15 | 3a67daa20b5 |
hex | 1f0304fb99d |
2131114310045 has 64 divisors (see below), whose sum is σ = 3210955776000. Its totient is φ = 1323942402816.
The previous prime is 2131114310041. The next prime is 2131114310069. The reversal of 2131114310045 is 5400134111312.
It is a cyclic number.
It is not a de Polignac number, because 2131114310045 - 22 = 2131114310041 is a prime.
It is a super-3 number, since 3×21311143100453 (a number of 38 digits) contains 333 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 2131114309996 and 2131114310023.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (2131114310041) by changing a digit.
It is a polite number, since it can be written in 63 ways as a sum of consecutive naturals, for example, 86236151 + ... + 86260859.
It is an arithmetic number, because the mean of its divisors is an integer number (50171184000).
Almost surely, 22131114310045 is an apocalyptic number.
It is an amenable number.
2131114310045 is a deficient number, since it is larger than the sum of its proper divisors (1079841465955).
2131114310045 is a wasteful number, since it uses less digits than its factorization.
2131114310045 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 30402.
The product of its (nonzero) digits is 1440, while the sum is 26.
Adding to 2131114310045 its reverse (5400134111312), we get a palindrome (7531248421357).
The spelling of 2131114310045 in words is "two trillion, one hundred thirty-one billion, one hundred fourteen million, three hundred ten thousand, forty-five".
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