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49105621153 = 180127265753
BaseRepresentation
bin101101101110111011…
…000101000010100001
311200202020220000111101
4231232323011002201
51301032014334103
634320411514401
73355611615151
oct555673050241
9150666800441
1049105621153
111990992aa86
129625458a01
134827693a84
14253ba38761
1514260c8a1d
hexb6eec50a1

49105621153 has 4 divisors (see below), whose sum is σ = 49132888708. Its totient is φ = 49078353600.

The previous prime is 49105621151. The next prime is 49105621171. The reversal of 49105621153 is 35112650194.

49105621153 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

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

It can be written as a sum of positive squares in 2 ways, for example, as 22682468449 + 26423152704 = 150607^2 + 162552^2 .

It is a cyclic number.

It is not a de Polignac number, because 49105621153 - 21 = 49105621151 is a prime.

It is a super-2 number, since 2×491056211532 (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 (49105621151) 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, 13631076 + ... + 13634677.

It is an arithmetic number, because the mean of its divisors is an integer number (12283222177).

Almost surely, 249105621153 is an apocalyptic number.

It is an amenable number.

49105621153 is a deficient number, since it is larger than the sum of its proper divisors (27267555).

49105621153 is a wasteful number, since it uses less digits than its factorization.

49105621153 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 27267554.

The product of its (nonzero) digits is 32400, while the sum is 37.

The spelling of 49105621153 in words is "forty-nine billion, one hundred five million, six hundred twenty-one thousand, one hundred fifty-three".

Divisors: 1 1801 27265753 49105621153