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123550032147 = 3331388132059
BaseRepresentation
bin111001100010000100…
…1111110000100010011
3102210220102011221002120
41303010021332010103
54011012302012042
6132431425545323
711632453642233
oct1630411760423
9383812157076
10123550032147
1148440854893
121bb407a5243
13b85c8208c9
145da0a09cc3
153331996dec
hex1cc427e113

123550032147 has 16 divisors (see below), whose sum is σ = 165278789760. Its totient is φ = 82094126400.

The previous prime is 123550032127. The next prime is 123550032161. The reversal of 123550032147 is 741230055321.

It is not a de Polignac number, because 123550032147 - 212 = 123550028051 is a prime.

It is a super-3 number, since 3×1235500321473 (a number of 34 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is not an unprimeable number, because it can be changed into a prime (123550032107) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 3837804 + ... + 3869862.

It is an arithmetic number, because the mean of its divisors is an integer number (10329924360).

Almost surely, 2123550032147 is an apocalyptic number.

123550032147 is a deficient number, since it is larger than the sum of its proper divisors (41728757613).

123550032147 is a wasteful number, since it uses less digits than its factorization.

123550032147 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 36274.

The product of its (nonzero) digits is 25200, while the sum is 33.

Adding to 123550032147 its reverse (741230055321), we get a palindrome (864780087468).

The spelling of 123550032147 in words is "one hundred twenty-three billion, five hundred fifty million, thirty-two thousand, one hundred forty-seven".

Divisors: 1 3 331 993 3881 11643 32059 96177 1284611 3853833 10611529 31834587 124420979 373262937 41183344049 123550032147