Base | Representation |
---|---|
bin | 1011111100100101001110… |
… | …0100101111101011001010 |
3 | 1201111201210110022121001011 |
4 | 2333021103210233223022 |
5 | 3210202311001220204 |
6 | 43534154112314134 |
7 | 2524000520626015 |
oct | 277112344575312 |
9 | 51451713277034 |
10 | 13135412132554 |
11 | 4204773789148 |
12 | 1581892b4294a |
13 | 743882bba53a |
14 | 335a861d437c |
15 | 17ba37710304 |
hex | bf25392faca |
13135412132554 has 4 divisors (see below), whose sum is σ = 19703118198834. Its totient is φ = 6567706066276.
The previous prime is 13135412132537. The next prime is 13135412132581. The reversal of 13135412132554 is 45523121453131.
It is a semiprime because it is the product of two primes.
It can be written as a sum of positive squares in only one way, i.e., 7720590174025 + 5414821958529 = 2778595^2 + 2326977^2 .
It is a super-3 number, since 3×131354121325543 (a number of 40 digits) contains 333 as substring.
It is an unprimeable number.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 3283853033137 + ... + 3283853033140.
Almost surely, 213135412132554 is an apocalyptic number.
13135412132554 is a deficient number, since it is larger than the sum of its proper divisors (6567706066280).
13135412132554 is an equidigital number, since it uses as much as digits as its factorization.
13135412132554 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 6567706066279.
The product of its digits is 216000, while the sum is 40.
Adding to 13135412132554 its reverse (45523121453131), we get a palindrome (58658533585685).
The spelling of 13135412132554 in words is "thirteen trillion, one hundred thirty-five billion, four hundred twelve million, one hundred thirty-two thousand, five hundred fifty-four".
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