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131032000113 = 3717934858207
BaseRepresentation
bin111101000001000011…
…1011010011001110001
3110112012211211021122020
41322002013122121301
54121323143000423
6140110102302053
712316044424020
oct1720207323161
9415184737566
10131032000113
115063016a093
122148a427929
13c4829371a1
1464b05868b7
15361d7664e3
hex1e821da671

131032000113 has 16 divisors (see below), whose sum is σ = 200783278080. Its totient is φ = 74457128016.

The previous prime is 131032000039. The next prime is 131032000147. The reversal of 131032000113 is 311000230131.

It is not a de Polignac number, because 131032000113 - 229 = 130495129201 is a prime.

It is a super-2 number, since 2×1310320001132 (a number of 23 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 131032000092 and 131032000101.

It is not an unprimeable number, because it can be changed into a prime (131032000193) by changing a digit.

It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 17425345 + ... + 17432862.

It is an arithmetic number, because the mean of its divisors is an integer number (12548954880).

Almost surely, 2131032000113 is an apocalyptic number.

It is an amenable number.

131032000113 is a deficient number, since it is larger than the sum of its proper divisors (69751277967).

131032000113 is a wasteful number, since it uses less digits than its factorization.

131032000113 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 34858396.

The product of its (nonzero) digits is 54, while the sum is 15.

Adding to 131032000113 its reverse (311000230131), we get a palindrome (442032230244).

The spelling of 131032000113 in words is "one hundred thirty-one billion, thirty-two million, one hundred thirteen", and thus it is an aban number.

Divisors: 1 3 7 21 179 537 1253 3759 34858207 104574621 244007449 732022347 6239619053 18718857159 43677333371 131032000113