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334011335433 = 31934671235281
BaseRepresentation
bin1001101110001001010…
…00001000001100001001
31011221010211020212212020
410313010220020030021
520433023400213213
6413235342035053
733063053335566
oct4670450101411
91157124225766
10334011335433
11119720766a86
1254897878a89
1325660283a69
141224821186d
158a4d54b923
hex4dc4a08309

334011335433 has 16 divisors (see below), whose sum is σ = 448614893376. Its totient is φ = 221045944320.

The previous prime is 334011335407. The next prime is 334011335461. The reversal of 334011335433 is 334533110433.

It is not a de Polignac number, because 334011335433 - 216 = 334011269897 is a prime.

It is a super-2 number, since 2×3340113354332 (a number of 24 digits) contains 22 as substring.

It is a Curzon number.

It is a junction number, because it is equal to n+sod(n) for n = 334011335394 and 334011335403.

It is not an unprimeable number, because it can be changed into a prime (334011335533) by changing a digit.

It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 347248 + ... + 888033.

It is an arithmetic number, because the mean of its divisors is an integer number (28038430836).

Almost surely, 2334011335433 is an apocalyptic number.

It is an amenable number.

334011335433 is a deficient number, since it is larger than the sum of its proper divisors (114603557943).

334011335433 is a wasteful number, since it uses less digits than its factorization.

334011335433 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 1235944.

The product of its (nonzero) digits is 58320, while the sum is 33.

Adding to 334011335433 its reverse (334533110433), we get a palindrome (668544445866).

The spelling of 334011335433 in words is "three hundred thirty-four billion, eleven million, three hundred thirty-five thousand, four hundred thirty-three".

Divisors: 1 3 193 467 579 1401 90131 270393 1235281 3705843 238409233 576876227 715227699 1730628681 111337111811 334011335433