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201312220012153 is a prime number
BaseRepresentation
bin101101110001011110100111…
…010111001101111001111001
3222101210021000202002110002021
4231301132213113031321321
5202341244411310342103
61552053320202234441
760255220323346615
oct5561364727157171
9871707022073067
10201312220012153
115916503aaa8538
121a6b37a71b5a21
138843878c90c33
14379d7c60dc345
1518418d5be3bbd
hexb717a75cde79

201312220012153 has 2 divisors, whose sum is σ = 201312220012154. Its totient is φ = 201312220012152.

The previous prime is 201312220012129. The next prime is 201312220012223. The reversal of 201312220012153 is 351210022213102.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 186664124850064 + 14648095162089 = 13662508^2 + 3827283^2 .

It is a cyclic number.

It is not a de Polignac number, because 201312220012153 - 25 = 201312220012121 is a prime.

It is not a weakly prime, because it can be changed into another prime (201312220012853) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 100656110006076 + 100656110006077.

It is an arithmetic number, because the mean of its divisors is an integer number (100656110006077).

Almost surely, 2201312220012153 is an apocalyptic number.

It is an amenable number.

201312220012153 is a deficient number, since it is larger than the sum of its proper divisors (1).

201312220012153 is an equidigital number, since it uses as much as digits as its factorization.

201312220012153 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 1440, while the sum is 25.

Adding to 201312220012153 its reverse (351210022213102), we get a palindrome (552522242225255).

The spelling of 201312220012153 in words is "two hundred one trillion, three hundred twelve billion, two hundred twenty million, twelve thousand, one hundred fifty-three".