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20130003311147 is a prime number
BaseRepresentation
bin1001001001110111000011…
…01100100101101000101011
32122021102001112121222022122
410210323201230211220223
510114302221321424042
6110451332412141455
74145226150646502
oct444734154455053
978242045558278
1020130003311147
1164610a5098954
1223113b546788b
13b30335a77712
144d8423d49839
1524d9631507d2
hex124ee1b25a2b

20130003311147 has 2 divisors, whose sum is σ = 20130003311148. Its totient is φ = 20130003311146.

The previous prime is 20130003311081. The next prime is 20130003311183. The reversal of 20130003311147 is 74111330003102.

It is a happy number.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 20130003311147 - 28 = 20130003310891 is a prime.

It is a super-3 number, since 3×201300033111473 (a number of 41 digits) contains 333 as substring.

It is not a weakly prime, because it can be changed into another prime (20130003311947) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 10065001655573 + 10065001655574.

It is an arithmetic number, because the mean of its divisors is an integer number (10065001655574).

Almost surely, 220130003311147 is an apocalyptic number.

20130003311147 is a deficient number, since it is larger than the sum of its proper divisors (1).

20130003311147 is an equidigital number, since it uses as much as digits as its factorization.

20130003311147 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 1512, while the sum is 26.

Adding to 20130003311147 its reverse (74111330003102), we get a palindrome (94241333314249).

The spelling of 20130003311147 in words is "twenty trillion, one hundred thirty billion, three million, three hundred eleven thousand, one hundred forty-seven".