Base | Representation |
---|---|
bin | 1001011010101111011011… |
… | …0111110100010000000111 |
3 | 1100122221002122211202201122 |
4 | 2112223312313310100013 |
5 | 2324124031312313224 |
6 | 34005005521544155 |
7 | 2116060606215326 |
oct | 226536667642007 |
9 | 40587078752648 |
10 | 10355013010439 |
11 | 33325999872a9 |
12 | 11b2a54aa505b |
13 | 5a162048c5c6 |
14 | 27b28468c0bd |
15 | 12e55711d15e |
hex | 96af6df4407 |
10355013010439 has 2 divisors, whose sum is σ = 10355013010440. Its totient is φ = 10355013010438.
The previous prime is 10355013010403. The next prime is 10355013010453. The reversal of 10355013010439 is 93401031055301.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 10355013010439 - 28 = 10355013010183 is a prime.
It is a super-3 number, since 3×103550130104393 (a number of 40 digits) contains 333 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 10355013010399 and 10355013010408.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (10355013010739) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5177506505219 + 5177506505220.
It is an arithmetic number, because the mean of its divisors is an integer number (5177506505220).
Almost surely, 210355013010439 is an apocalyptic number.
10355013010439 is a deficient number, since it is larger than the sum of its proper divisors (1).
10355013010439 is an equidigital number, since it uses as much as digits as its factorization.
10355013010439 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 24300, while the sum is 35.
The spelling of 10355013010439 in words is "ten trillion, three hundred fifty-five billion, thirteen million, ten thousand, four hundred thirty-nine".
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