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1101131210303 is a prime number
BaseRepresentation
bin10000000001100000100…
…010001101111000111111
310220021012201122012112122
4100001200202031320333
5121020104042212203
62201504103153155
7142361025566525
oct20014042157077
93807181565478
101101131210303
11394a94953043
121594a67457bb
137cab4265a97
143b41b80c315
151d999e4e538
hex1006088de3f

1101131210303 has 2 divisors, whose sum is σ = 1101131210304. Its totient is φ = 1101131210302.

The previous prime is 1101131210293. The next prime is 1101131210317. The reversal of 1101131210303 is 3030121311011.

It is a weak prime.

It is a cyclic number.

It is not a de Polignac number, because 1101131210303 - 24 = 1101131210287 is a prime.

It is a super-2 number, since 2×11011312103032 (a number of 25 digits) contains 22 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (1101131210323) 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, 550565605151 + 550565605152.

It is an arithmetic number, because the mean of its divisors is an integer number (550565605152).

Almost surely, 21101131210303 is an apocalyptic number.

1101131210303 is a deficient number, since it is larger than the sum of its proper divisors (1).

1101131210303 is an equidigital number, since it uses as much as digits as its factorization.

1101131210303 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 54, while the sum is 17.

Adding to 1101131210303 its reverse (3030121311011), we get a palindrome (4131252521314).

The spelling of 1101131210303 in words is "one trillion, one hundred one billion, one hundred thirty-one million, two hundred ten thousand, three hundred three".