Search a number
-
+
201520034 = 2100760017
BaseRepresentation
bin11000000001011…
…11001110100010
3112001012021111222
430000233032202
5403042120114
631555135042
74664614652
oct1400571642
9461167458
10201520034
11a3830a91
12575a4482
133299a131
141ca9a362
1512a5998e
hexc02f3a2

201520034 has 4 divisors (see below), whose sum is σ = 302280054. Its totient is φ = 100760016.

The previous prime is 201520031. The next prime is 201520049. The reversal of 201520034 is 430025102.

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

It can be written as a sum of positive squares in only one way, i.e., 201214225 + 305809 = 14185^2 + 553^2 .

It is a super-2 number, since 2×2015200342 = 81220648206722312, which contains 22 as substring.

It is a Curzon number.

It is a junction number, because it is equal to n+sod(n) for n = 201519994 and 201520021.

It is not an unprimeable number, because it can be changed into a prime (201520031) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 50380007 + ... + 50380010.

Almost surely, 2201520034 is an apocalyptic number.

201520034 is a deficient number, since it is larger than the sum of its proper divisors (100760020).

201520034 is a wasteful number, since it uses less digits than its factorization.

201520034 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 100760019.

The product of its (nonzero) digits is 240, while the sum is 17.

The square root of 201520034 is about 14195.7752165917. The cubic root of 201520034 is about 586.2813455628.

Adding to 201520034 its reverse (430025102), we get a palindrome (631545136).

The spelling of 201520034 in words is "two hundred one million, five hundred twenty thousand, thirty-four".

Divisors: 1 2 100760017 201520034