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31004230433 = 1119331814591
BaseRepresentation
bin11100110111111111…
…110000001100100001
32222000201221002112212
4130313333300030201
51001444040333213
622124304131505
72145212416241
oct346777601441
988021832485
1031004230433
1112170097240
126013306b95
132c014539a7
141701977b21
15c16d939a8
hex737ff0321

31004230433 has 16 divisors (see below), whose sum is σ = 34016140032. Its totient is φ = 28024704000.

The previous prime is 31004230423. The next prime is 31004230459. The reversal of 31004230433 is 33403240013.

It is a cyclic number.

It is not a de Polignac number, because 31004230433 - 24 = 31004230417 is a prime.

It is a super-2 number, since 2×310042304332 (a number of 22 digits) contains 22 as substring.

It is a Duffinian number.

It is a Curzon number.

It is a junction number, because it is equal to n+sod(n) for n = 31004230399 and 31004230408.

It is not an unprimeable number, because it can be changed into a prime (31004230423) 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, 6750968 + ... + 6755558.

It is an arithmetic number, because the mean of its divisors is an integer number (2126008752).

Almost surely, 231004230433 is an apocalyptic number.

It is an amenable number.

31004230433 is a deficient number, since it is larger than the sum of its proper divisors (3011909599).

31004230433 is a wasteful number, since it uses less digits than its factorization.

31004230433 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 7976.

The product of its (nonzero) digits is 2592, while the sum is 23.

Adding to 31004230433 its reverse (33403240013), we get a palindrome (64407470446).

The spelling of 31004230433 in words is "thirty-one billion, four million, two hundred thirty thousand, four hundred thirty-three".

Divisors: 1 11 193 2123 3181 4591 34991 50501 613933 886063 6753263 9746693 14603971 160643681 2818566403 31004230433