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113413314312 = 23321575184921
BaseRepresentation
bin110100110011111110…
…1011101011100001000
3101211201222011022100200
41221213331131130020
53324232302024222
6124033520532200
711123326160064
oct1514775353410
9354658138320
10113413314312
114410994a731
1219b91a81060
13a90570500c
1456bc658aa4
152e3baceeac
hex1a67f5d708

113413314312 has 24 divisors (see below), whose sum is σ = 307161059790. Its totient is φ = 37804438080.

The previous prime is 113413314281. The next prime is 113413314331. The reversal of 113413314312 is 213413314311.

113413314312 is a `hidden beast` number, since 1 + 1 + 34 + 1 + 3 + 314 + 312 = 666.

113413314312 is digitally balanced in base 3, because in such base it contains all the possibile digits an equal number of times.

It can be written as a sum of positive squares in only one way, i.e., 112569644196 + 843670116 = 335514^2 + 29046^2 .

It is a tau number, because it is divible by the number of its divisors (24).

It is a nude number because it is divisible by every one of its digits.

It is an unprimeable number.

It is a polite number, since it can be written in 5 ways as a sum of consecutive naturals, for example, 787592389 + ... + 787592532.

Almost surely, 2113413314312 is an apocalyptic number.

113413314312 is a gapful number since it is divisible by the number (12) formed by its first and last digit.

It is an amenable number.

113413314312 is an abundant number, since it is smaller than the sum of its proper divisors (193747745478).

It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.

113413314312 is a wasteful number, since it uses less digits than its factorization.

113413314312 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 1575184933 (or 1575184926 counting only the distinct ones).

The product of its digits is 2592, while the sum is 27.

Adding to 113413314312 its reverse (213413314311), we get a palindrome (326826628623).

Subtracting 113413314312 from its reverse (213413314311), we obtain a palindrome (99999999999).

The spelling of 113413314312 in words is "one hundred thirteen billion, four hundred thirteen million, three hundred fourteen thousand, three hundred twelve".

Divisors: 1 2 3 4 6 8 9 12 18 24 36 72 1575184921 3150369842 4725554763 6300739684 9451109526 12601479368 14176664289 18902219052 28353328578 37804438104 56706657156 113413314312