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201520131 = 367173377
BaseRepresentation
bin11000000001011…
…11010000000011
3112001012021122120
430000233100003
5403042121011
631555135323
74664615151
oct1400572003
9461167576
10201520131
11a383106a
12575a4543
133299a1a7
141ca9a3d1
1512a59a06
hexc02f403

201520131 has 4 divisors (see below), whose sum is σ = 268693512. Its totient is φ = 134346752.

The previous prime is 201520117. The next prime is 201520133. The reversal of 201520131 is 131025102.

201520131 is nontrivially palindromic in base 2.

It is a semiprime because it is the product of two primes.

It is a cyclic number.

It is not a de Polignac number, because 201520131 - 27 = 201520003 is a prime.

It is a super-2 number, since 2×2015201312 = 81220726396514322, which contains 22 as substring.

It is a self number, because there is not a number n which added to its sum of digits gives 201520131.

It is not an unprimeable number, because it can be changed into a prime (201520133) 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, 33586686 + ... + 33586691.

It is an arithmetic number, because the mean of its divisors is an integer number (67173378).

Almost surely, 2201520131 is an apocalyptic number.

201520131 is a deficient number, since it is larger than the sum of its proper divisors (67173381).

201520131 is an equidigital number, since it uses as much as digits as its factorization.

201520131 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 67173380.

The product of its (nonzero) digits is 60, while the sum is 15.

The square root of 201520131 is about 14195.7786331008. The cubic root of 201520131 is about 586.2814396300.

Adding to 201520131 its reverse (131025102), we get a palindrome (332545233).

The spelling of 201520131 in words is "two hundred one million, five hundred twenty thousand, one hundred thirty-one".

Divisors: 1 3 67173377 201520131