Search a number
-
+
210411412142131 is a prime number
BaseRepresentation
bin101111110101111000111001…
…011110100011100000110011
31000121000010122002101120100101
4233311320321132203200303
5210034340033002022011
62023301400204333231
762214502641331414
oct5765707136434063
91017003562346311
10210411412142131
1161052a84815a92
121b723184244217
139053932183255
1439d5d7682b80b
15194d43c72b2c1
hexbf5e397a3833

210411412142131 has 2 divisors, whose sum is σ = 210411412142132. Its totient is φ = 210411412142130.

The previous prime is 210411412142021. The next prime is 210411412142143. The reversal of 210411412142131 is 131241214114012.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 210411412142131 - 215 = 210411412109363 is a prime.

It is a super-2 number, since 2×2104114121421312 (a number of 29 digits) contains 22 as substring.

It is a self number, because there is not a number n which added to its sum of digits gives 210411412142131.

It is not a weakly prime, because it can be changed into another prime (210411412142161) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 105205706071065 + 105205706071066.

It is an arithmetic number, because the mean of its divisors is an integer number (105205706071066).

Almost surely, 2210411412142131 is an apocalyptic number.

210411412142131 is a deficient number, since it is larger than the sum of its proper divisors (1).

210411412142131 is an equidigital number, since it uses as much as digits as its factorization.

210411412142131 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 1536, while the sum is 28.

Adding to 210411412142131 its reverse (131241214114012), we get a palindrome (341652626256143).

The spelling of 210411412142131 in words is "two hundred ten trillion, four hundred eleven billion, four hundred twelve million, one hundred forty-two thousand, one hundred thirty-one".