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1330220320034 = 2665110160017
BaseRepresentation
bin10011010110110111010…
…011110100010100100010
311201011112100012202212112
4103112313103310110202
5133243242400220114
62455032245555322
7165051046105604
oct23266723642442
94634470182775
101330220320034
114731637679a1
12195980115b42
139859300c808
1448550d85174
152490706e13e
hex135b74f4522

1330220320034 has 4 divisors (see below), whose sum is σ = 1995330480054. Its totient is φ = 665110160016.

The previous prime is 1330220320027. The next prime is 1330220320049. The reversal of 1330220320034 is 4300230220331.

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., 1304860713025 + 25359607009 = 1142305^2 + 159247^2 .

It is a super-2 number, since 2×13302203200342 (a number of 25 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 1330220319991 and 1330220320009.

It is an unprimeable number.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 332555080007 + ... + 332555080010.

Almost surely, 21330220320034 is an apocalyptic number.

1330220320034 is a deficient number, since it is larger than the sum of its proper divisors (665110160020).

1330220320034 is an equidigital number, since it uses as much as digits as its factorization.

1330220320034 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 665110160019.

The product of its (nonzero) digits is 2592, while the sum is 23.

Adding to 1330220320034 its reverse (4300230220331), we get a palindrome (5630450540365).

The spelling of 1330220320034 in words is "one trillion, three hundred thirty billion, two hundred twenty million, three hundred twenty thousand, thirty-four".

Divisors: 1 2 665110160017 1330220320034