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130123324011314 = 265061662005657
BaseRepresentation
bin11101100101100010110010…
…100000010011001100110010
3122001201200001011220011120102
4131211202302200103030302
5114023420031421330224
61140425455123312402
736260046162413114
oct3545426240231462
9561650034804512
10130123324011314
1138508aa0107237
1212716931b1b702
13577b76630aa4a
14241c006b713b4
151009c162964ae
hex7658b2813332

130123324011314 has 4 divisors (see below), whose sum is σ = 195184986016974. Its totient is φ = 65061662005656.

The previous prime is 130123324011313. The next prime is 130123324011347. The reversal of 130123324011314 is 413110423321031.

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., 75168605220289 + 54954718791025 = 8669983^2 + 7413145^2 .

It is not an unprimeable number, because it can be changed into a prime (130123324011313) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 32530831002827 + ... + 32530831002830.

Almost surely, 2130123324011314 is an apocalyptic number.

130123324011314 is a deficient number, since it is larger than the sum of its proper divisors (65061662005660).

130123324011314 is an equidigital number, since it uses as much as digits as its factorization.

130123324011314 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 65061662005659.

The product of its (nonzero) digits is 5184, while the sum is 29.

Adding to 130123324011314 its reverse (413110423321031), we get a palindrome (543233747332345).

The spelling of 130123324011314 in words is "one hundred thirty trillion, one hundred twenty-three billion, three hundred twenty-four million, eleven thousand, three hundred fourteen".

Divisors: 1 2 65061662005657 130123324011314