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13302440143142 = 26651220071571
BaseRepresentation
bin1100000110010011011100…
…1110000010100100100110
31202002200220220222200010112
43001210313032002210212
53220421344134040032
644143020122253022
72542032603403013
oct301446716024446
952080826880115
1013302440143142
114269595737717
1215aa128357172
137565511b96b7
1433dbad27670a
1518106119b1b2
hexc1937382926

13302440143142 has 4 divisors (see below), whose sum is σ = 19953660214716. Its totient is φ = 6651220071570.

The previous prime is 13302440143127. The next prime is 13302440143151. The reversal of 13302440143142 is 24134104420331.

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

It is a junction number, because it is equal to n+sod(n) for n = 13302440143099 and 13302440143108.

It is a congruent number.

It is an unprimeable number.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 3325610035784 + ... + 3325610035787.

It is an arithmetic number, because the mean of its divisors is an integer number (4988415053679).

Almost surely, 213302440143142 is an apocalyptic number.

13302440143142 is a deficient number, since it is larger than the sum of its proper divisors (6651220071574).

13302440143142 is an equidigital number, since it uses as much as digits as its factorization.

13302440143142 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 6651220071573.

The product of its (nonzero) digits is 27648, while the sum is 32.

Adding to 13302440143142 its reverse (24134104420331), we get a palindrome (37436544563473).

The spelling of 13302440143142 in words is "thirteen trillion, three hundred two billion, four hundred forty million, one hundred forty-three thousand, one hundred forty-two".

Divisors: 1 2 6651220071571 13302440143142