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201343541557 = 1910597028503
BaseRepresentation
bin1011101110000100000…
…0101010000100110101
3201020200222101200000211
42323201000222010311
511244322421312212
6232255042505421
720355323602534
oct2734100520465
9636628350024
10201343541557
11784311a8924
1233031665271
1315ca9790992
149a606d831b
1553863b59a7
hex2ee102a135

201343541557 has 4 divisors (see below), whose sum is σ = 211940570080. Its totient is φ = 190746513036.

The previous prime is 201343541537. The next prime is 201343541569. The reversal of 201343541557 is 755145343102.

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 201343541557 - 215 = 201343508789 is a prime.

It is a Duffinian number.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (201343541537) 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, 5298514233 + ... + 5298514270.

It is an arithmetic number, because the mean of its divisors is an integer number (52985142520).

Almost surely, 2201343541557 is an apocalyptic number.

It is an amenable number.

201343541557 is a deficient number, since it is larger than the sum of its proper divisors (10597028523).

201343541557 is a wasteful number, since it uses less digits than its factorization.

201343541557 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 10597028522.

The product of its (nonzero) digits is 252000, while the sum is 40.

Adding to 201343541557 its reverse (755145343102), we get a palindrome (956488884659).

The spelling of 201343541557 in words is "two hundred one billion, three hundred forty-three million, five hundred forty-one thousand, five hundred fifty-seven".

Divisors: 1 19 10597028503 201343541557