Base | Representation |
---|---|
bin | 10011100111000011000100… |
… | …10000011101100000000111 |
3 | 12122200112022101110010200011 |
4 | 21303201202100131200013 |
5 | 21123012122223243101 |
6 | 231414253215331051 |
7 | 12040354055136211 |
oct | 1163414220354007 |
9 | 178615271403604 |
10 | 43123120134151 |
11 | 1281645a109932 |
12 | 4a05671820a87 |
13 | 1b0a653839245 |
14 | a91254c735b1 |
15 | 4ebae82da751 |
hex | 27386241d807 |
43123120134151 has 2 divisors, whose sum is σ = 43123120134152. Its totient is φ = 43123120134150.
The previous prime is 43123120134053. The next prime is 43123120134277. The reversal of 43123120134151 is 15143102132134.
It is a happy number.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 43123120134151 - 225 = 43123086579719 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (43123120134851) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 21561560067075 + 21561560067076.
It is an arithmetic number, because the mean of its divisors is an integer number (21561560067076).
Almost surely, 243123120134151 is an apocalyptic number.
43123120134151 is a deficient number, since it is larger than the sum of its proper divisors (1).
43123120134151 is an equidigital number, since it uses as much as digits as its factorization.
43123120134151 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 8640, while the sum is 31.
Adding to 43123120134151 its reverse (15143102132134), we get a palindrome (58266222266285).
The spelling of 43123120134151 in words is "forty-three trillion, one hundred twenty-three billion, one hundred twenty million, one hundred thirty-four thousand, one hundred fifty-one".
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