Base | Representation |
---|---|
bin | 11110110100100101111101… |
… | …010011000110011111110011 |
3 | 122202221222121101122222222222 |
4 | 132310211331103012133303 |
5 | 120231420244114112103 |
6 | 1200145211124151255 |
7 | 40360365161022110 |
oct | 3664457523063763 |
9 | 582858541588888 |
10 | 135555564988403 |
11 | 3a2128795673aa |
12 | 132536b053852b |
13 | 5a83ab1453281 |
14 | 2568cd25cb907 |
15 | 10a11a0aed638 |
hex | 7b497d4c67f3 |
135555564988403 has 16 divisors (see below), whose sum is σ = 154969760784384. Its totient is φ = 116153650909440.
The previous prime is 135555564988373. The next prime is 135555564988411. The reversal of 135555564988403 is 304889465555531.
It is a cyclic number.
It is not a de Polignac number, because 135555564988403 - 214 = 135555564972019 is a prime.
It is a super-2 number, since 2×1355555649884032 (a number of 29 digits) contains 22 as substring.
It is not an unprimeable number, because it can be changed into a prime (135555564988703) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (29) of ones.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 622485326 + ... + 622703052.
It is an arithmetic number, because the mean of its divisors is an integer number (9685610049024).
Almost surely, 2135555564988403 is an apocalyptic number.
135555564988403 is a deficient number, since it is larger than the sum of its proper divisors (19414195795981).
135555564988403 is a wasteful number, since it uses less digits than its factorization.
135555564988403 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 245522.
The product of its (nonzero) digits is 1555200000, while the sum is 71.
The spelling of 135555564988403 in words is "one hundred thirty-five trillion, five hundred fifty-five billion, five hundred sixty-four million, nine hundred eighty-eight thousand, four hundred three".
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