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4333621247981 is a prime number
BaseRepresentation
bin111111000011111111111…
…101000111111111101101
3120100021211111221002101222
4333003333331013333231
51032000224014413411
613114501011113125
7625044113553554
oct77037775077755
916307744832358
104333621247981
111420973243123
1259ba74ba27a5
132558742cb9cc
1410da6980ad9b
1577ada1504db
hex3f0fff47fed

4333621247981 has 2 divisors, whose sum is σ = 4333621247982. Its totient is φ = 4333621247980.

The previous prime is 4333621247959. The next prime is 4333621248007. The reversal of 4333621247981 is 1897421263334.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 3290432742025 + 1043188505956 = 1813955^2 + 1021366^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-4333621247981 is a prime.

It is a super-2 number, since 2×43336212479812 (a number of 26 digits) contains 22 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (4333621247281) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2166810623990 + 2166810623991.

It is an arithmetic number, because the mean of its divisors is an integer number (2166810623991).

Almost surely, 24333621247981 is an apocalyptic number.

It is an amenable number.

4333621247981 is a deficient number, since it is larger than the sum of its proper divisors (1).

4333621247981 is an equidigital number, since it uses as much as digits as its factorization.

4333621247981 is an evil number, because the sum of its binary digits is even.

The product of its digits is 5225472, while the sum is 53.

The spelling of 4333621247981 in words is "four trillion, three hundred thirty-three billion, six hundred twenty-one million, two hundred forty-seven thousand, nine hundred eighty-one".