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1130330540434 = 2565165270217
BaseRepresentation
bin10000011100101100111…
…100110100100110010010
311000001120200022212100011
4100130230330310212102
5122004404104243214
62223133334145134
7144443434032064
oct20345474644622
94001520285304
101130330540434
113a6409138a37
121630954b21aa
13827887a2114
143c9cb795734
151e60861cac4
hex1072cf34992

1130330540434 has 4 divisors (see below), whose sum is σ = 1695495810654. Its totient is φ = 565165270216.

The previous prime is 1130330540359. The next prime is 1130330540483. The reversal of 1130330540434 is 4340450330311.

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., 1004110214809 + 126220325625 = 1002053^2 + 355275^2 .

It is a junction number, because it is equal to n+sod(n) for n = 1130330540396 and 1130330540405.

It is an unprimeable number.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 282582635107 + ... + 282582635110.

Almost surely, 21130330540434 is an apocalyptic number.

1130330540434 is a deficient number, since it is larger than the sum of its proper divisors (565165270220).

1130330540434 is an equidigital number, since it uses as much as digits as its factorization.

1130330540434 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 565165270219.

The product of its (nonzero) digits is 25920, while the sum is 31.

Adding to 1130330540434 its reverse (4340450330311), we get a palindrome (5470780870745).

The spelling of 1130330540434 in words is "one trillion, one hundred thirty billion, three hundred thirty million, five hundred forty thousand, four hundred thirty-four".

Divisors: 1 2 565165270217 1130330540434