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143113105353 = 3113294210819
BaseRepresentation
bin1000010101001000110…
…1001010111111001001
3111200101210120000110110
42011102031022333021
54321043423332403
6145424550304533
713224322216131
oct2052215127711
9450353500413
10143113105353
115576a687289
12238a0358149
131065982817a
146cd8c77ac1
153ac91a8103
hex215234afc9

143113105353 has 8 divisors (see below), whose sum is σ = 190834362400. Its totient is φ = 95400292608.

The previous prime is 143113105309. The next prime is 143113105361. The reversal of 143113105353 is 353501311341.

It is a sphenic number, since it is the product of 3 distinct primes.

It is not a de Polignac number, because 143113105353 - 213 = 143113097161 is a prime.

It is a Duffinian number.

It is a Curzon number.

It is a self number, because there is not a number n which added to its sum of digits gives 143113105353.

It is not an unprimeable number, because it can be changed into a prime (143113105153) by changing a digit.

It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 2071423 + ... + 2139396.

It is an arithmetic number, because the mean of its divisors is an integer number (23854295300).

Almost surely, 2143113105353 is an apocalyptic number.

It is an amenable number.

143113105353 is a deficient number, since it is larger than the sum of its proper divisors (47721257047).

143113105353 is a wasteful number, since it uses less digits than its factorization.

143113105353 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 4222151.

The product of its (nonzero) digits is 8100, while the sum is 30.

Adding to 143113105353 its reverse (353501311341), we get a palindrome (496614416694).

The spelling of 143113105353 in words is "one hundred forty-three billion, one hundred thirteen million, one hundred five thousand, three hundred fifty-three".

Divisors: 1 3 11329 33987 4210819 12632457 47704368451 143113105353