Base | Representation |
---|---|
bin | 11101111000101101011100… |
… | …111110011100110100110001 |
3 | 122020101120202112120220100221 |
4 | 131320231130332130310301 |
5 | 114212010012133044403 |
6 | 1143314523440413041 |
7 | 36454154510034226 |
oct | 3570553476346461 |
9 | 566346675526327 |
10 | 131440444034353 |
11 | 389766410150a3 |
12 | 128aa056917181 |
13 | 5845a1c22cb1c |
14 | 2465a73d1d94d |
15 | 102e1031a0dbd |
hex | 778b5cf9cd31 |
131440444034353 has 2 divisors, whose sum is σ = 131440444034354. Its totient is φ = 131440444034352.
The previous prime is 131440444034329. The next prime is 131440444034399. The reversal of 131440444034353 is 353430444044131.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 126207092361249 + 5233351673104 = 11234193^2 + 2287652^2 .
It is a cyclic number.
It is not a de Polignac number, because 131440444034353 - 25 = 131440444034321 is a prime.
It is not a weakly prime, because it can be changed into another prime (131440444034323) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65720222017176 + 65720222017177.
It is an arithmetic number, because the mean of its divisors is an integer number (65720222017177).
Almost surely, 2131440444034353 is an apocalyptic number.
It is an amenable number.
131440444034353 is a deficient number, since it is larger than the sum of its proper divisors (1).
131440444034353 is an equidigital number, since it uses as much as digits as its factorization.
131440444034353 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 1658880, while the sum is 43.
Adding to 131440444034353 its reverse (353430444044131), we get a palindrome (484870888078484).
The spelling of 131440444034353 in words is "one hundred thirty-one trillion, four hundred forty billion, four hundred forty-four million, thirty-four thousand, three hundred fifty-three".
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