Base | Representation |
---|---|
bin | 11101110100000011001110… |
… | …111011010001001010010101 |
3 | 122012020222020111111221020121 |
4 | 131310003032323101022111 |
5 | 114141233214203134243 |
6 | 1142511445324231541 |
7 | 36422060420235352 |
oct | 3564031673211225 |
9 | 565228214457217 |
10 | 131120233255573 |
11 | 38862862414641 |
12 | 12857b9099b5b1 |
13 | 582177b167123 |
14 | 2454378909629 |
15 | 1025b114dd1ed |
hex | 7740ceed1295 |
131120233255573 has 2 divisors, whose sum is σ = 131120233255574. Its totient is φ = 131120233255572.
The previous prime is 131120233255523. The next prime is 131120233255577. The reversal of 131120233255573 is 375552332021131.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 102640058486449 + 28480174769124 = 10131143^2 + 5336682^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-131120233255573 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (131120233255577) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65560116627786 + 65560116627787.
It is an arithmetic number, because the mean of its divisors is an integer number (65560116627787).
Almost surely, 2131120233255573 is an apocalyptic number.
It is an amenable number.
131120233255573 is a deficient number, since it is larger than the sum of its proper divisors (1).
131120233255573 is an equidigital number, since it uses as much as digits as its factorization.
131120233255573 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 567000, while the sum is 43.
The spelling of 131120233255573 in words is "one hundred thirty-one trillion, one hundred twenty billion, two hundred thirty-three million, two hundred fifty-five thousand, five hundred seventy-three".
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