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11314304433 = 3219122354101
BaseRepresentation
bin10101000100110001…
…01001010110110001
31002012111212012212100
422202120221112301
5141132430220213
65110412451013
7550244000205
oct124230512661
932174765770
1011314304433
1148866a2423
122239179a69
1310b40932c3
14794928b05
15463476773
hex2a26295b1

11314304433 has 24 divisors (see below), whose sum is σ = 17217420480. Its totient is φ = 7139901600.

The previous prime is 11314304413. The next prime is 11314304459. The reversal of 11314304433 is 33440341311.

11314304433 is a `hidden beast` number, since 1 + 1 + 314 + 304 + 43 + 3 = 666.

It is not a de Polignac number, because 11314304433 - 210 = 11314303409 is a prime.

It is a Curzon number.

It is a junction number, because it is equal to n+sod(n) for n = 11314304397 and 11314304406.

It is not an unprimeable number, because it can be changed into a prime (11314304413) by changing a digit.

It is a polite number, since it can be written in 23 ways as a sum of consecutive naturals, for example, 182083 + ... + 236183.

It is an arithmetic number, because the mean of its divisors is an integer number (717392520).

Almost surely, 211314304433 is an apocalyptic number.

It is an amenable number.

11314304433 is a deficient number, since it is larger than the sum of its proper divisors (5903116047).

11314304433 is a wasteful number, since it uses less digits than its factorization.

11314304433 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 55349 (or 55346 counting only the distinct ones).

The product of its (nonzero) digits is 5184, while the sum is 27.

Adding to 11314304433 its reverse (33440341311), we get a palindrome (44754645744).

The spelling of 11314304433 in words is "eleven billion, three hundred fourteen million, three hundred four thousand, four hundred thirty-three".

Divisors: 1 3 9 19 57 171 1223 3669 11007 23237 54101 69711 162303 209133 486909 1027919 3083757 9251271 66165523 198496569 595489707 1257144937 3771434811 11314304433