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131142101243851 is a prime number
BaseRepresentation
bin11101110100010111100110…
…010111000001011111001011
3122012100001200111222211200101
4131310113212113001133023
5114142113010404300401
61142525503303150231
736423463344124255
oct3564274627013713
9565301614884611
10131142101243851
1138871064314925
1212860274445377
135823854847a69
142455450d24cd5
151026491242101
hex7745e65c17cb

131142101243851 has 2 divisors, whose sum is σ = 131142101243852. Its totient is φ = 131142101243850.

The previous prime is 131142101243807. The next prime is 131142101243893. The reversal of 131142101243851 is 158342101241131.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 131142101243851 - 219 = 131142100719563 is a prime.

It is a self number, because there is not a number n which added to its sum of digits gives 131142101243851.

It is not a weakly prime, because it can be changed into another prime (131142101243801) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65571050621925 + 65571050621926.

It is an arithmetic number, because the mean of its divisors is an integer number (65571050621926).

Almost surely, 2131142101243851 is an apocalyptic number.

131142101243851 is a deficient number, since it is larger than the sum of its proper divisors (1).

131142101243851 is an equidigital number, since it uses as much as digits as its factorization.

131142101243851 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 23040, while the sum is 37.

Adding to 131142101243851 its reverse (158342101241131), we get a palindrome (289484202484982).

The spelling of 131142101243851 in words is "one hundred thirty-one trillion, one hundred forty-two billion, one hundred one million, two hundred forty-three thousand, eight hundred fifty-one".