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530433250041 = 31551111399077
BaseRepresentation
bin1111011100000000100…
…10010001001011111001
31212201010210201222221110
413232000102101023321
532142311403000131
61043402213230533
753215426143012
oct7560022211371
91781123658843
10530433250041
11194a56817712
1286975032449
133b0341c6836
141b95d00a009
15dbe77dc646
hex7b804912f9

530433250041 has 8 divisors (see below), whose sum is σ = 707289991744. Its totient is φ = 353599337520.

The previous prime is 530433250027. The next prime is 530433250051. The reversal of 530433250041 is 140052334035.

It is a sphenic number, since it is the product of 3 distinct primes.

It is not a de Polignac number, because 530433250041 - 218 = 530432987897 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 530433249990 and 530433250008.

It is not an unprimeable number, because it can be changed into a prime (530433250001) by changing a digit.

It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 5653006 + ... + 5746071.

It is an arithmetic number, because the mean of its divisors is an integer number (88411248968).

Almost surely, 2530433250041 is an apocalyptic number.

It is an amenable number.

530433250041 is a deficient number, since it is larger than the sum of its proper divisors (176856741703).

530433250041 is a wasteful number, since it uses less digits than its factorization.

530433250041 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 11414591.

The product of its (nonzero) digits is 21600, while the sum is 30.

Adding to 530433250041 its reverse (140052334035), we get a palindrome (670485584076).

The spelling of 530433250041 in words is "five hundred thirty billion, four hundred thirty-three million, two hundred fifty thousand, forty-one".

Divisors: 1 3 15511 46533 11399077 34197231 176811083347 530433250041