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134020141102 = 267010070551
BaseRepresentation
bin111110011010000111…
…0010001000000101110
3110210221002112120110201
41330310032101000232
54143433124003402
6141322404424114
712453100255231
oct1746416210056
9423832476421
10134020141102
115192396a6a1
1221b830a803a
13c83aa28953
1466b5383418
153745c66687
hex1f3439102e

134020141102 has 4 divisors (see below), whose sum is σ = 201030211656. Its totient is φ = 67010070550.

The previous prime is 134020141097. The next prime is 134020141159. The reversal of 134020141102 is 201141020431.

It is a semiprime because it is the product of two primes.

It is a super-2 number, since 2×1340201411022 (a number of 23 digits) contains 22 as substring.

It is a self number, because there is not a number n which added to its sum of digits gives 134020141102.

It is a congruent number.

It is an unprimeable number.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 33505035274 + ... + 33505035277.

It is an arithmetic number, because the mean of its divisors is an integer number (50257552914).

Almost surely, 2134020141102 is an apocalyptic number.

134020141102 is a deficient number, since it is larger than the sum of its proper divisors (67010070554).

134020141102 is an equidigital number, since it uses as much as digits as its factorization.

134020141102 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 67010070553.

The product of its (nonzero) digits is 192, while the sum is 19.

Adding to 134020141102 its reverse (201141020431), we get a palindrome (335161161533).

The spelling of 134020141102 in words is "one hundred thirty-four billion, twenty million, one hundred forty-one thousand, one hundred two".

Divisors: 1 2 67010070551 134020141102