Base | Representation |
---|---|
bin | 11101110111010011110110… |
… | …011010101110010100101100 |
3 | 122020001101102111122110122112 |
4 | 131313103312122232110230 |
5 | 114203420442413043200 |
6 | 1143202412544103152 |
7 | 36444210364141526 |
oct | 3567236632562454 |
9 | 566041374573575 |
10 | 131344234112300 |
11 | 38939860023720 |
12 | 12893486304ab8 |
13 | 58399289c2571 |
14 | 24611483aaa16 |
15 | 102b871a72535 |
hex | 7774f66ae52c |
131344234112300 has 72 divisors (see below), whose sum is σ = 311064662097840. Its totient is φ = 47740489171200.
The previous prime is 131344234112293. The next prime is 131344234112309. The reversal of 131344234112300 is 3211432443131.
It is a super-2 number, since 2×1313442341123002 (a number of 29 digits) contains 22 as substring.
It is not an unprimeable number, because it can be changed into a prime (131344234112309) by changing a digit.
It is a polite number, since it can be written in 23 ways as a sum of consecutive naturals, for example, 23816099 + ... + 28807898.
Almost surely, 2131344234112300 is an apocalyptic number.
131344234112300 is a gapful number since it is divisible by the number (10) formed by its first and last digit.
It is an amenable number.
131344234112300 is an abundant number, since it is smaller than the sum of its proper divisors (179720427985540).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
131344234112300 is a wasteful number, since it uses less digits than its factorization.
131344234112300 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 52626291 (or 52626284 counting only the distinct ones).
The product of its (nonzero) digits is 20736, while the sum is 32.
Adding to 131344234112300 its reverse (3211432443131), we get a palindrome (134555666555431).
The spelling of 131344234112300 in words is "one hundred thirty-one trillion, three hundred forty-four billion, two hundred thirty-four million, one hundred twelve thousand, three hundred".
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