Base | Representation |
---|---|
bin | 1011011001111101000001… |
… | …1011010100100101010111 |
3 | 1122101212020110011010002010 |
4 | 2312133100123110211113 |
5 | 3120430424112011101 |
6 | 42401010141115303 |
7 | 2433010303166340 |
oct | 266372033244527 |
9 | 48355213133063 |
10 | 12540506360151 |
11 | 3aa5442a01200 |
12 | 14a6526258b33 |
13 | 6cc745790bb3 |
14 | 314d6c73d0c7 |
15 | 16b319c0d2d6 |
hex | b67d06d4957 |
12540506360151 has 48 divisors (see below), whose sum is σ = 21006686591744. Its totient is φ = 6513865776000.
The previous prime is 12540506360143. The next prime is 12540506360231. The reversal of 12540506360151 is 15106360504521.
It is not a de Polignac number, because 12540506360151 - 23 = 12540506360143 is a prime.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (12540506360101) by changing a digit.
It is a polite number, since it can be written in 47 ways as a sum of consecutive naturals, for example, 24447216 + ... + 24954906.
Almost surely, 212540506360151 is an apocalyptic number.
12540506360151 is a gapful number since it is divisible by the number (11) formed by its first and last digit.
12540506360151 is a deficient number, since it is larger than the sum of its proper divisors (8466180231593).
12540506360151 is a wasteful number, since it uses less digits than its factorization.
12540506360151 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 517444 (or 517433 counting only the distinct ones).
The product of its (nonzero) digits is 108000, while the sum is 39.
Adding to 12540506360151 its reverse (15106360504521), we get a palindrome (27646866864672).
The spelling of 12540506360151 in words is "twelve trillion, five hundred forty billion, five hundred six million, three hundred sixty thousand, one hundred fifty-one".
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