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350054005433 = 11611094786147
BaseRepresentation
bin1010001100000001101…
…10000011101010111001
31020110112212100221220012
411012000312003222321
521213402311133213
6424451304111305
734201443644336
oct5060066035271
91213485327805
10350054005433
111255033a6440
1257a14465535
1327018b29bc2
1412d2aacd98d
15918bb704a8
hex5180d83ab9

350054005433 has 16 divisors (see below), whose sum is σ = 391698352320. Its totient is φ = 310142260800.

The previous prime is 350054005429. The next prime is 350054005439. The reversal of 350054005433 is 334500450053.

It is a cyclic number.

It is not a de Polignac number, because 350054005433 - 22 = 350054005429 is a prime.

It is a self number, because there is not a number n which added to its sum of digits gives 350054005433.

It is not an unprimeable number, because it can be changed into a prime (350054005439) by changing a digit.

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 in 15 ways as a sum of consecutive naturals, for example, 2319935 + ... + 2466212.

It is an arithmetic number, because the mean of its divisors is an integer number (24481147020).

Almost surely, 2350054005433 is an apocalyptic number.

It is an amenable number.

350054005433 is a deficient number, since it is larger than the sum of its proper divisors (41644346887).

350054005433 is a wasteful number, since it uses less digits than its factorization.

350054005433 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 4786328.

The product of its (nonzero) digits is 54000, while the sum is 32.

Adding to 350054005433 its reverse (334500450053), we get a palindrome (684554455486).

The spelling of 350054005433 in words is "three hundred fifty billion, fifty-four million, five thousand, four hundred thirty-three".

Divisors: 1 11 61 109 671 1199 6649 73139 4786147 52647617 291954967 521690023 3211504637 5738590253 31823091403 350054005433