Base | Representation |
---|---|
bin | 110001010111100001011011… |
… | …0010001000010000110110011 |
3 | 2002221112002000222110010021121 |
4 | 1202233002312101002012303 |
5 | 423404102422420210141 |
6 | 4135315432025203111 |
7 | 160315623323441014 |
oct | 14257026621020663 |
9 | 2087462028403247 |
10 | 434241431413171 |
11 | 1163a9836422448 |
12 | 40852a94930497 |
13 | 1583ba39583b95 |
14 | 793357c75100b |
15 | 353093981b1d1 |
hex | 18af0b64421b3 |
434241431413171 has 2 divisors, whose sum is σ = 434241431413172. Its totient is φ = 434241431413170.
The previous prime is 434241431413157. The next prime is 434241431413261. The reversal of 434241431413171 is 171314134142434.
It is a weak prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-434241431413171 is a prime.
It is a super-2 number, since 2×4342414314131712 (a number of 30 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (434241431411171) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 217120715706585 + 217120715706586.
It is an arithmetic number, because the mean of its divisors is an integer number (217120715706586).
Almost surely, 2434241431413171 is an apocalyptic number.
434241431413171 is a deficient number, since it is larger than the sum of its proper divisors (1).
434241431413171 is an equidigital number, since it uses as much as digits as its factorization.
434241431413171 is an evil number, because the sum of its binary digits is even.
The product of its digits is 387072, while the sum is 43.
The spelling of 434241431413171 in words is "four hundred thirty-four trillion, two hundred forty-one billion, four hundred thirty-one million, four hundred thirteen thousand, one hundred seventy-one".
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