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141421200433 = 634392229247
BaseRepresentation
bin1000001110110101011…
…1000100010000110001
3111112000212221100020021
42003231113010100301
54304112311403213
6144545031043441
713134356240542
oct2035527042061
9445025840207
10141421200433
1154a81643057
12234a9810581
131044a1543b6
146bb829c4c9
153a2a8a2d8d
hex20ed5c4431

141421200433 has 4 divisors (see below), whose sum is σ = 141423493120. Its totient is φ = 141418907748.

The previous prime is 141421200379. The next prime is 141421200449. The reversal of 141421200433 is 334002124141.

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 a de Polignac number, because none of the positive numbers 2k-141421200433 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 141421200398 and 141421200407.

It is not an unprimeable number, because it can be changed into a prime (141421200133) 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, 1051185 + ... + 1178062.

It is an arithmetic number, because the mean of its divisors is an integer number (35355873280).

Almost surely, 2141421200433 is an apocalyptic number.

It is an amenable number.

141421200433 is a deficient number, since it is larger than the sum of its proper divisors (2292687).

141421200433 is an equidigital number, since it uses as much as digits as its factorization.

141421200433 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 2292686.

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

Adding to 141421200433 its reverse (334002124141), we get a palindrome (475423324574).

The spelling of 141421200433 in words is "one hundred forty-one billion, four hundred twenty-one million, two hundred thousand, four hundred thirty-three".

Divisors: 1 63439 2229247 141421200433