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11331011013002 = 25665505506501
BaseRepresentation
bin1010010011100011010011…
…1000101110100110001010
31111010020022122021101112122
42210320310320232212022
52441121402304404002
640033221122211242
72246432005314422
oct244706470564612
944106278241478
1011331011013002
11367950035527a
1213300387b3b22
13642681c0c8a5
142b25d10d9482
15149b2b5eb6a2
hexa4e34e2e98a

11331011013002 has 4 divisors (see below), whose sum is σ = 16996516519506. Its totient is φ = 5665505506500.

The previous prime is 11331011012963. The next prime is 11331011013011. The reversal of 11331011013002 is 20031011013311.

It is a semiprime because it is the product of two primes.

It can be written as a sum of positive squares in only one way, i.e., 10444718685241 + 886292327761 = 3231829^2 + 941431^2 .

It is a super-2 number, since 2×113310110130022 (a number of 27 digits) contains 22 as substring.

It is an unprimeable number.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2832752753249 + ... + 2832752753252.

Almost surely, 211331011013002 is an apocalyptic number.

11331011013002 is a deficient number, since it is larger than the sum of its proper divisors (5665505506504).

11331011013002 is an equidigital number, since it uses as much as digits as its factorization.

11331011013002 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 5665505506503.

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

Adding to 11331011013002 its reverse (20031011013311), we get a palindrome (31362022026313).

The spelling of 11331011013002 in words is "eleven trillion, three hundred thirty-one billion, eleven million, thirteen thousand, two".

Divisors: 1 2 5665505506501 11331011013002