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10011320143 = 11314395227
BaseRepresentation
bin10010101001011100…
…01001111101001111
3221211201001202102221
421110232021331033
5131000344221033
64333225200211
7503042550166
oct112456117517
927751052387
1010011320143
114278147000
121b34930067
13c37151b96
146ad86c3dd
153d8d9712d
hex254b89f4f

10011320143 has 16 divisors (see below), whose sum is σ = 11021460480. Its totient is φ = 9093135480.

The previous prime is 10011320113. The next prime is 10011320159. The reversal of 10011320143 is 34102311001.

It is not a de Polignac number, because 10011320143 - 29 = 10011319631 is a prime.

It is a super-4 number, since 4×100113201434 (a number of 41 digits) contains 4444 as substring.

It is a Duffinian number.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (10011320113) by changing a digit.

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 in 15 ways as a sum of consecutive naturals, for example, 1912696 + ... + 1917922.

It is an arithmetic number, because the mean of its divisors is an integer number (688841280).

Almost surely, 210011320143 is an apocalyptic number.

10011320143 is a deficient number, since it is larger than the sum of its proper divisors (1010140337).

10011320143 is an equidigital number, since it uses as much as digits as its factorization.

10011320143 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 6699 (or 6677 counting only the distinct ones).

The product of its (nonzero) digits is 72, while the sum is 16.

Adding to 10011320143 its reverse (34102311001), we get a palindrome (44113631144).

The spelling of 10011320143 in words is "ten billion, eleven million, three hundred twenty thousand, one hundred forty-three".

Divisors: 1 11 121 1331 1439 5227 15829 57497 174119 632467 1915309 6957137 7521653 82738183 910120013 10011320143