Base | Representation |
---|---|
bin | 10001011111001000001… |
… | …011001000101101100111 |
3 | 11020212200100110002222120 |
4 | 101133020023020231213 |
5 | 124141441441104210 |
6 | 2320010552401023 |
7 | 152550053331345 |
oct | 21371013105547 |
9 | 4225610402876 |
10 | 1201654238055 |
11 | 423689543687 |
12 | 174a7b6a6173 |
13 | 89414230bab |
14 | 422360da995 |
15 | 213d0050270 |
hex | 117c82c8b67 |
1201654238055 has 32 divisors (see below), whose sum is σ = 2036357708544. Its totient is φ = 603001360896.
The previous prime is 1201654238017. The next prime is 1201654238077. The reversal of 1201654238055 is 5508324561021.
It is not a de Polignac number, because 1201654238055 - 214 = 1201654221671 is a prime.
It is a super-2 number, since 2×12016542380552 (a number of 25 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 1201654237998 and 1201654238016.
It is a congruent number.
It is an unprimeable number.
It is a polite number, since it can be written in 31 ways as a sum of consecutive naturals, for example, 138312 + ... + 1556418.
It is an arithmetic number, because the mean of its divisors is an integer number (63636178392).
Almost surely, 21201654238055 is an apocalyptic number.
1201654238055 is a gapful number since it is divisible by the number (15) formed by its first and last digit.
1201654238055 is a deficient number, since it is larger than the sum of its proper divisors (834703470489).
1201654238055 is a wasteful number, since it uses less digits than its factorization.
1201654238055 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 1421455.
The product of its (nonzero) digits is 288000, while the sum is 42.
Adding to 1201654238055 its reverse (5508324561021), we get a palindrome (6709978799076).
The spelling of 1201654238055 in words is "one trillion, two hundred one billion, six hundred fifty-four million, two hundred thirty-eight thousand, fifty-five".
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