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10344314041 = 15168505391
BaseRepresentation
bin10011010001001000…
…11011010010111001
3222200220122121012021
421220210123102321
5132141121022131
64430242314441
7514225134655
oct115044332271
928626577167
1010344314041
114429106276
122008358a21
13c8b1325a9
14701b8da65
15408221e11
hex26891b4b9

10344314041 has 4 divisors (see below), whose sum is σ = 10412819584. Its totient is φ = 10275808500.

The previous prime is 10344314021. The next prime is 10344314069. The reversal of 10344314041 is 14041344301.

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 10344314041 - 211 = 10344311993 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 10344313997 and 10344314015.

It is not an unprimeable number, because it can be changed into a prime (10344314011) 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, 34252545 + ... + 34252846.

It is an arithmetic number, because the mean of its divisors is an integer number (2603204896).

Almost surely, 210344314041 is an apocalyptic number.

It is an amenable number.

10344314041 is a deficient number, since it is larger than the sum of its proper divisors (68505543).

10344314041 is an equidigital number, since it uses as much as digits as its factorization.

10344314041 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 68505542.

The product of its (nonzero) digits is 2304, while the sum is 25.

Adding to 10344314041 its reverse (14041344301), we get a palindrome (24385658342).

The spelling of 10344314041 in words is "ten billion, three hundred forty-four million, three hundred fourteen thousand, forty-one".

Divisors: 1 151 68505391 10344314041