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13210114113 = 327209684351
BaseRepresentation
bin11000100110110001…
…00101100001000001
31021002122011022211200
430103120211201001
5204023242122423
610022454331413
7645234055140
oct142330454101
937078138750
1013210114113
115669846289
122688059b69
131326a8a791
148d4621757
15524b07d43
hex313625841

13210114113 has 12 divisors (see below), whose sum is σ = 21807172608. Its totient is φ = 7548636600.

The previous prime is 13210114081. The next prime is 13210114123. The reversal of 13210114113 is 31141101231.

It is a happy number.

It is not a de Polignac number, because 13210114113 - 25 = 13210114081 is a prime.

It is a super-2 number, since 2×132101141132 (a number of 21 digits) contains 22 as substring.

It is a Curzon number.

It is not an unprimeable number, because it can be changed into a prime (13210114123) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (13) of ones.

It is a polite number, since it can be written in 11 ways as a sum of consecutive naturals, for example, 104842113 + ... + 104842238.

It is an arithmetic number, because the mean of its divisors is an integer number (1817264384).

Almost surely, 213210114113 is an apocalyptic number.

It is an amenable number.

13210114113 is a deficient number, since it is larger than the sum of its proper divisors (8597058495).

13210114113 is a wasteful number, since it uses less digits than its factorization.

13210114113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 209684364 (or 209684361 counting only the distinct ones).

The product of its (nonzero) digits is 72, while the sum is 18.

Adding to 13210114113 its reverse (31141101231), we get a palindrome (44351215344).

The spelling of 13210114113 in words is "thirteen billion, two hundred ten million, one hundred fourteen thousand, one hundred thirteen".

Divisors: 1 3 7 9 21 63 209684351 629053053 1467790457 1887159159 4403371371 13210114113