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14020123153 = 14239852511
BaseRepresentation
bin11010000111010101…
…00001101000010001
31100012002100001100111
431003222201220101
5212203122420103
610235111521321
71004301036421
oct150352415021
940162301314
1014020123153
115a44aa3043
122873389241
13142583637a
14970034481
15570caa86d
hex343aa1a11

14020123153 has 4 divisors (see below), whose sum is σ = 14029977088. Its totient is φ = 14010269220.

The previous prime is 14020123129. The next prime is 14020123157. The reversal of 14020123153 is 35132102041.

It is a happy number.

It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.

It is a cyclic number.

It is not a de Polignac number, because 14020123153 - 25 = 14020123121 is a prime.

It is a super-2 number, since 2×140201231532 (a number of 21 digits) contains 22 as substring.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (14020123157) by changing a digit.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 4924833 + ... + 4927678.

It is an arithmetic number, because the mean of its divisors is an integer number (3507494272).

Almost surely, 214020123153 is an apocalyptic number.

It is an amenable number.

14020123153 is a deficient number, since it is larger than the sum of its proper divisors (9853935).

14020123153 is an equidigital number, since it uses as much as digits as its factorization.

14020123153 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 9853934.

The product of its (nonzero) digits is 720, while the sum is 22.

Adding to 14020123153 its reverse (35132102041), we get a palindrome (49152225194).

The spelling of 14020123153 in words is "fourteen billion, twenty million, one hundred twenty-three thousand, one hundred fifty-three".

Divisors: 1 1423 9852511 14020123153