Base | Representation |
---|---|
bin | 11110110100011010100011… |
… | …101011100101110001101111 |
3 | 122202220211000022011101222211 |
4 | 132310122203223211301233 |
5 | 120231220211423242203 |
6 | 1200135424330151251 |
7 | 40356452632463554 |
oct | 3664324353456157 |
9 | 582824008141884 |
10 | 135543324040303 |
11 | 3a2086678a4973 |
12 | 13251254b82527 |
13 | 5a828b03c820a |
14 | 25684909aca2b |
15 | 10a0bd6135c6d |
hex | 7b46a3ae5c6f |
135543324040303 has 2 divisors, whose sum is σ = 135543324040304. Its totient is φ = 135543324040302.
The previous prime is 135543324040297. The next prime is 135543324040307. The reversal of 135543324040303 is 303040423345531.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 135543324040303 - 237 = 135405885086831 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (135543324040307) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 67771662020151 + 67771662020152.
It is an arithmetic number, because the mean of its divisors is an integer number (67771662020152).
Almost surely, 2135543324040303 is an apocalyptic number.
135543324040303 is a deficient number, since it is larger than the sum of its proper divisors (1).
135543324040303 is an equidigital number, since it uses as much as digits as its factorization.
135543324040303 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 777600, while the sum is 40.
Adding to 135543324040303 its reverse (303040423345531), we get a palindrome (438583747385834).
The spelling of 135543324040303 in words is "one hundred thirty-five trillion, five hundred forty-three billion, three hundred twenty-four million, forty thousand, three hundred three".
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