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43303121847131 is a prime number
BaseRepresentation
bin10011101100010010010110…
…01100100000001101011011
312200022201221000102120200112
421312021023030200031123
521133434243143102011
6232033102440554535
712056355504352346
oct1166111314401533
9180281830376615
1043303121847131
1112885829410aa3
124a34527a82a4b
131b2160b930393
14a99c4cd9745d
15501630c3158b
hex27624b32035b

43303121847131 has 2 divisors, whose sum is σ = 43303121847132. Its totient is φ = 43303121847130.

The previous prime is 43303121847119. The next prime is 43303121847157. The reversal of 43303121847131 is 13174812130334.

It is a weak prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-43303121847131 is a prime.

It is a super-2 number, since 2×433031218471312 (a number of 28 digits) contains 22 as substring.

It is a Sophie Germain prime.

It is not a weakly prime, because it can be changed into another prime (43303121247131) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 21651560923565 + 21651560923566.

It is an arithmetic number, because the mean of its divisors is an integer number (21651560923566).

Almost surely, 243303121847131 is an apocalyptic number.

43303121847131 is a deficient number, since it is larger than the sum of its proper divisors (1).

43303121847131 is an equidigital number, since it uses as much as digits as its factorization.

43303121847131 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 145152, while the sum is 41.

Adding to 43303121847131 its reverse (13174812130334), we get a palindrome (56477933977465).

The spelling of 43303121847131 in words is "forty-three trillion, three hundred three billion, one hundred twenty-one million, eight hundred forty-seven thousand, one hundred thirty-one".