Base | Representation |
---|---|
bin | 1011111100110111001011… |
… | …1000111111100011001110 |
3 | 1201112012020101200201001122 |
4 | 2333031302320333203032 |
5 | 3210242144010021304 |
6 | 43540312305440542 |
7 | 2524231132242452 |
oct | 277156270774316 |
9 | 51465211621048 |
10 | 13140232501454 |
11 | 42068177556a3 |
12 | 15827b9340152 |
13 | 744170774126 |
14 | 335dc2496662 |
15 | 17bc1a9d29be |
hex | bf372e3f8ce |
13140232501454 has 4 divisors (see below), whose sum is σ = 19710348752184. Its totient is φ = 6570116250726.
The previous prime is 13140232501421. The next prime is 13140232501457. The reversal of 13140232501454 is 45410523204131.
It is a semiprime because it is the product of two primes.
It is a super-3 number, since 3×131402325014543 (a number of 40 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (13140232501457) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 3285058125362 + ... + 3285058125365.
It is an arithmetic number, because the mean of its divisors is an integer number (4927587188046).
Almost surely, 213140232501454 is an apocalyptic number.
13140232501454 is a deficient number, since it is larger than the sum of its proper divisors (6570116250730).
13140232501454 is an equidigital number, since it uses as much as digits as its factorization.
13140232501454 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 6570116250729.
The product of its (nonzero) digits is 57600, while the sum is 35.
Adding to 13140232501454 its reverse (45410523204131), we get a palindrome (58550755705585).
The spelling of 13140232501454 in words is "thirteen trillion, one hundred forty billion, two hundred thirty-two million, five hundred one thousand, four hundred fifty-four".
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