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57117201403 = 120427474289
BaseRepresentation
bin110101001100011100…
…110101001111111011
312110102121002110111211
4311030130311033323
51413434000421103
642123403503551
74061263126033
oct651434651773
9173377073454
1057117201403
1122250197889
12b0a0527bb7
13550342cac2
142a9ba788c3
15174461036d
hexd4c7353fb

57117201403 has 4 divisors (see below), whose sum is σ = 57117796120. Its totient is φ = 57116606688.

The previous prime is 57117201389. The next prime is 57117201431. The reversal of 57117201403 is 30410271175.

It is a semiprime because it is the product of two primes, and also a brilliant number, because the two primes have the same length.

It is a cyclic number.

It is not a de Polignac number, because 57117201403 - 221 = 57115104251 is a prime.

It is a super-2 number, since 2×571172014032 (a number of 22 digits) contains 22 as substring.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (57117201433) 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, 116718 + ... + 357571.

It is an arithmetic number, because the mean of its divisors is an integer number (14279449030).

Almost surely, 257117201403 is an apocalyptic number.

57117201403 is a deficient number, since it is larger than the sum of its proper divisors (594717).

57117201403 is a wasteful number, since it uses less digits than its factorization.

57117201403 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 594716.

The product of its (nonzero) digits is 5880, while the sum is 31.

Adding to 57117201403 its reverse (30410271175), we get a palindrome (87527472578).

The spelling of 57117201403 in words is "fifty-seven billion, one hundred seventeen million, two hundred one thousand, four hundred three".

Divisors: 1 120427 474289 57117201403