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1002012001153 is a prime number
BaseRepresentation
bin11101001010011001001…
…00011100001110000001
310112210100221100200012021
432211030210130032001
5112404110033014103
62044152402401441
7132251535323422
oct16451444341601
93483327320167
101002012001153
11356a5071a169
12142243960281
1373649085577
14366d7747649
151b0e81e9bbd
hexe94c91c381

1002012001153 has 2 divisors, whose sum is σ = 1002012001154. Its totient is φ = 1002012001152.

The previous prime is 1002012001139. The next prime is 1002012001241. The reversal of 1002012001153 is 3511002102001.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 837027371664 + 164984629489 = 914892^2 + 406183^2 .

It is a cyclic number.

It is not a de Polignac number, because 1002012001153 - 29 = 1002012000641 is a prime.

It is not a weakly prime, because it can be changed into another prime (1002012001253) 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, 501006000576 + 501006000577.

It is an arithmetic number, because the mean of its divisors is an integer number (501006000577).

Almost surely, 21002012001153 is an apocalyptic number.

It is an amenable number.

1002012001153 is a deficient number, since it is larger than the sum of its proper divisors (1).

1002012001153 is an equidigital number, since it uses as much as digits as its factorization.

1002012001153 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 60, while the sum is 16.

Adding to 1002012001153 its reverse (3511002102001), we get a palindrome (4513014103154).

The spelling of 1002012001153 in words is "one trillion, two billion, twelve million, one thousand, one hundred fifty-three".