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113762138700493 is a prime number
BaseRepresentation
bin11001110111011101001111…
…110111011011101011001101
3112220210112222001020111120201
4121313131033313123223031
5104402334330001403433
61041541334404413501
732651020564035403
oct3167351767335315
9486715861214521
10113762138700493
1133280285583998
1210913a460a7291
134b62953b1650a
14201418ca73473
15d243333c0a7d
hex67774fddbacd

113762138700493 has 2 divisors, whose sum is σ = 113762138700494. Its totient is φ = 113762138700492.

The previous prime is 113762138700491. The next prime is 113762138700521. The reversal of 113762138700493 is 394007831267311.

It is a happy number.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 113112860515209 + 649278185284 = 10635453^2 + 805778^2 .

It is a cyclic number.

It is not a de Polignac number, because 113762138700493 - 21 = 113762138700491 is a prime.

Together with 113762138700491, it forms a pair of twin primes.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (113762138700491) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 56881069350246 + 56881069350247.

It is an arithmetic number, because the mean of its divisors is an integer number (56881069350247).

Almost surely, 2113762138700493 is an apocalyptic number.

It is an amenable number.

113762138700493 is a deficient number, since it is larger than the sum of its proper divisors (1).

113762138700493 is an equidigital number, since it uses as much as digits as its factorization.

113762138700493 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 4572288, while the sum is 55.

The spelling of 113762138700493 in words is "one hundred thirteen trillion, seven hundred sixty-two billion, one hundred thirty-eight million, seven hundred thousand, four hundred ninety-three".