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11031235035 = 352925359161
BaseRepresentation
bin10100100011000001…
…10100100111011011
31001110210012210122120
422101200310213123
5140042444010120
65022341352323
7540235644042
oct122140644733
931423183576
1011031235035
114750932177
12217a4086a3
13106a542812
147690c1559
154486ae140
hex2918349db

11031235035 has 16 divisors (see below), whose sum is σ = 18258596640. Its totient is φ = 5680451840.

The previous prime is 11031235013. The next prime is 11031235051. The reversal of 11031235035 is 53053213011.

It is not a de Polignac number, because 11031235035 - 25 = 11031235003 is a prime.

It is a super-2 number, since 2×110312350352 (a number of 21 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 11031234996 and 11031235014.

It is an unprimeable number.

It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 12679146 + ... + 12680015.

It is an arithmetic number, because the mean of its divisors is an integer number (1141162290).

Almost surely, 211031235035 is an apocalyptic number.

11031235035 is a gapful number since it is divisible by the number (15) formed by its first and last digit.

11031235035 is a deficient number, since it is larger than the sum of its proper divisors (7227361605).

11031235035 is a wasteful number, since it uses less digits than its factorization.

11031235035 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 25359198.

The product of its (nonzero) digits is 1350, while the sum is 24.

Adding to 11031235035 its reverse (53053213011), we get a palindrome (64084448046).

The spelling of 11031235035 in words is "eleven billion, thirty-one million, two hundred thirty-five thousand, thirty-five".

Divisors: 1 3 5 15 29 87 145 435 25359161 76077483 126795805 380387415 735415669 2206247007 3677078345 11031235035