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130034523221147 is a prime number
BaseRepresentation
bin11101100100010000000101…
…100100001010110010011011
3122001102011211111112122002002
4131210100011210022302123
5114020441200421034042
61140321011342231215
736250453464224126
oct3544200544126233
9561364744478062
10130034523221147
1138484371402256
121270168a88550b
135773283a943ac
1424179c12393bd
15100776a34b132
hex76440590ac9b

130034523221147 has 2 divisors, whose sum is σ = 130034523221148. Its totient is φ = 130034523221146.

The previous prime is 130034523221129. The next prime is 130034523221171. The reversal of 130034523221147 is 741122325430031.

It is a weak prime.

It is an emirp because it is prime and its reverse (741122325430031) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 130034523221147 - 222 = 130034519026843 is a prime.

It is a self number, because there is not a number n which added to its sum of digits gives 130034523221147.

It is not a weakly prime, because it can be changed into another prime (130034513221147) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65017261610573 + 65017261610574.

It is an arithmetic number, because the mean of its divisors is an integer number (65017261610574).

Almost surely, 2130034523221147 is an apocalyptic number.

130034523221147 is a deficient number, since it is larger than the sum of its proper divisors (1).

130034523221147 is an equidigital number, since it uses as much as digits as its factorization.

130034523221147 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 120960, while the sum is 38.

Adding to 130034523221147 its reverse (741122325430031), we get a palindrome (871156848651178).

The spelling of 130034523221147 in words is "one hundred thirty trillion, thirty-four billion, five hundred twenty-three million, two hundred twenty-one thousand, one hundred forty-seven".