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57350113 = 193018427
BaseRepresentation
bin1101101011000…
…1011111100001
310222220200120021
43122301133201
5104140200423
65405113441
71264316332
oct332613741
9128820507
1057350113
112a410a97
1217258881
13bb5caa2
14788c289
15507c95d
hex36b17e1

57350113 has 4 divisors (see below), whose sum is σ = 60368560. Its totient is φ = 54331668.

The previous prime is 57350077. The next prime is 57350119. The reversal of 57350113 is 31105375.

It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.

It is a cyclic number.

It is not a de Polignac number, because 57350113 - 221 = 55252961 is a prime.

It is a super-2 number, since 2×573501132 = 6578070922225538, which contains 22 as substring.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (57350119) by changing a digit.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 1509195 + ... + 1509232.

It is an arithmetic number, because the mean of its divisors is an integer number (15092140).

Almost surely, 257350113 is an apocalyptic number.

It is an amenable number.

57350113 is a deficient number, since it is larger than the sum of its proper divisors (3018447).

57350113 is a wasteful number, since it uses less digits than its factorization.

57350113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 3018446.

The product of its (nonzero) digits is 1575, while the sum is 25.

The square root of 57350113 is about 7572.9857387955. The cubic root of 57350113 is about 385.6364645338.

Adding to 57350113 its reverse (31105375), we get a palindrome (88455488).

The spelling of 57350113 in words is "fifty-seven million, three hundred fifty thousand, one hundred thirteen".

Divisors: 1 19 3018427 57350113