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74623544034113 is a prime number
BaseRepresentation
bin10000111101111010100101…
…110101011000001101000001
3100210012221102002102101022102
4100331322211311120031001
534240113004233042423
6422413325522120145
721501241662340064
oct2075724565301501
9323187362371272
1074623544034113
11218607381a2618
128452661ab0055
133283c7c150969
14145db27cb28db
158961e2793c28
hex43dea5d58341

74623544034113 has 2 divisors, whose sum is σ = 74623544034114. Its totient is φ = 74623544034112.

The previous prime is 74623544034019. The next prime is 74623544034139. The reversal of 74623544034113 is 31143044532647.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 74623544034064 + 49 = 8638492^2 + 7^2 .

It is a cyclic number.

It is not a de Polignac number, because 74623544034113 - 210 = 74623544033089 is a prime.

It is a super-3 number, since 3×746235440341133 (a number of 43 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is not a weakly prime, because it can be changed into another prime (74623544034713) 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, 37311772017056 + 37311772017057.

It is an arithmetic number, because the mean of its divisors is an integer number (37311772017057).

Almost surely, 274623544034113 is an apocalyptic number.

It is an amenable number.

74623544034113 is a deficient number, since it is larger than the sum of its proper divisors (1).

74623544034113 is an equidigital number, since it uses as much as digits as its factorization.

74623544034113 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 2903040, while the sum is 47.

The spelling of 74623544034113 in words is "seventy-four trillion, six hundred twenty-three billion, five hundred forty-four million, thirty-four thousand, one hundred thirteen".