Base | Representation |
---|---|
bin | 111001010100110001… |
… | …0010101000110101011 |
3 | 102202202021211100102021 |
4 | 1302221202111012223 |
5 | 4004103434330011 |
6 | 132315233445311 |
7 | 11615421626260 |
oct | 1625142250653 |
9 | 382667740367 |
10 | 123103433131 |
11 | 48231751398 |
12 | 1ba37110837 |
13 | b7bb144c33 |
14 | 5d5b595267 |
15 | 330767b471 |
hex | 1ca98951ab |
123103433131 has 4 divisors (see below), whose sum is σ = 140689637872. Its totient is φ = 105517228392.
The previous prime is 123103433113. The next prime is 123103433137. The reversal of 123103433131 is 131334301321.
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is not a de Polignac number, because 123103433131 - 231 = 120955949483 is a prime.
It is a super-2 number, since 2×1231034331312 (a number of 23 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 123103433096 and 123103433105.
It is not an unprimeable number, because it can be changed into a prime (123103433137) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 8793102360 + ... + 8793102373.
It is an arithmetic number, because the mean of its divisors is an integer number (35172409468).
Almost surely, 2123103433131 is an apocalyptic number.
123103433131 is a deficient number, since it is larger than the sum of its proper divisors (17586204741).
123103433131 is an equidigital number, since it uses as much as digits as its factorization.
123103433131 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 17586204740.
The product of its (nonzero) digits is 1944, while the sum is 25.
Adding to 123103433131 its reverse (131334301321), we get a palindrome (254437734452).
The spelling of 123103433131 in words is "one hundred twenty-three billion, one hundred three million, four hundred thirty-three thousand, one hundred thirty-one".
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