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132994021143817 is a prime number
BaseRepresentation
bin11110001111010100010101…
…011000110011010100001001
3122102220002211000000012102001
4132033110111120303110021
5114412433223403100232
61150504335421511001
740004333552422141
oct3617242530632411
9572802730005361
10132994021143817
11394154a1874849
1212abb170712461
135929389510262
1424bad32dcb921
15105972de7e6e7
hex78f515633509

132994021143817 has 2 divisors, whose sum is σ = 132994021143818. Its totient is φ = 132994021143816.

The previous prime is 132994021143793. The next prime is 132994021143833. The reversal of 132994021143817 is 718341120499231.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 129778666497936 + 3215354645881 = 11392044^2 + 1793141^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-132994021143817 is a prime.

It is not a weakly prime, because it can be changed into another prime (132994021143917) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 66497010571908 + 66497010571909.

It is an arithmetic number, because the mean of its divisors is an integer number (66497010571909).

Almost surely, 2132994021143817 is an apocalyptic number.

It is an amenable number.

132994021143817 is a deficient number, since it is larger than the sum of its proper divisors (1).

132994021143817 is an equidigital number, since it uses as much as digits as its factorization.

132994021143817 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 2612736, while the sum is 55.

The spelling of 132994021143817 in words is "one hundred thirty-two trillion, nine hundred ninety-four billion, twenty-one million, one hundred forty-three thousand, eight hundred seventeen".