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1152012554113 is a prime number
BaseRepresentation
bin10000110000111001010…
…011001000111110000001
311002010112201212021120211
4100300321103020332001
5122333310203212423
62241121031034121
7146141642100001
oct20607123107601
94063481767524
101152012554113
11404625065803
1216732680b941
1384832799605
143da87200001
151ee76d8810d
hex10c394c8f81

1152012554113 has 2 divisors, whose sum is σ = 1152012554114. Its totient is φ = 1152012554112.

The previous prime is 1152012553973. The next prime is 1152012554159. The reversal of 1152012554113 is 3114552102511.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 802049207184 + 349963346929 = 895572^2 + 591577^2 .

It is a cyclic number.

It is not a de Polignac number, because 1152012554113 - 225 = 1151978999681 is a prime.

It is not a weakly prime, because it can be changed into another prime (1152012554213) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 576006277056 + 576006277057.

It is an arithmetic number, because the mean of its divisors is an integer number (576006277057).

Almost surely, 21152012554113 is an apocalyptic number.

It is an amenable number.

1152012554113 is a deficient number, since it is larger than the sum of its proper divisors (1).

1152012554113 is an equidigital number, since it uses as much as digits as its factorization.

1152012554113 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 6000, while the sum is 31.

Adding to 1152012554113 its reverse (3114552102511), we get a palindrome (4266564656624).

The spelling of 1152012554113 in words is "one trillion, one hundred fifty-two billion, twelve million, five hundred fifty-four thousand, one hundred thirteen".