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21213430540153 = 158144313413971
BaseRepresentation
bin1001101001011001000101…
…11111011110011101111001
32210002222120110002202011021
410310230202333132131321
510240030101204241103
6113041152045502441
74316422404312445
oct464544277363571
983088513082137
1021213430540153
116839624563795
12246737ab71a21
13bab558215807
14534a42385825
1526bc23bbabbd
hex134b22fde779

21213430540153 has 4 divisors (see below), whose sum is σ = 21213445535568. Its totient is φ = 21213415544740.

The previous prime is 21213430540111. The next prime is 21213430540229. The reversal of 21213430540153 is 35104503431212.

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 21213430540153 - 237 = 21075991586681 is a prime.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (21213430540553) 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, 5125543 + ... + 8288428.

It is an arithmetic number, because the mean of its divisors is an integer number (5303361383892).

Almost surely, 221213430540153 is an apocalyptic number.

It is an amenable number.

21213430540153 is a deficient number, since it is larger than the sum of its proper divisors (14995415).

21213430540153 is a wasteful number, since it uses less digits than its factorization.

21213430540153 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 14995414.

The product of its (nonzero) digits is 43200, while the sum is 34.

Adding to 21213430540153 its reverse (35104503431212), we get a palindrome (56317933971365).

The spelling of 21213430540153 in words is "twenty-one trillion, two hundred thirteen billion, four hundred thirty million, five hundred forty thousand, one hundred fifty-three".

Divisors: 1 1581443 13413971 21213430540153