Base | Representation |
---|---|
bin | 111101101010110110… |
… | …0000000111011110011 |
3 | 110122211120010112122201 |
4 | 1323111230000323303 |
5 | 4132211112312201 |
6 | 140501150244031 |
7 | 12365561211424 |
oct | 1732554007363 |
9 | 418746115581 |
10 | 132434104051 |
11 | 5118a665577 |
12 | 217bbab6617 |
13 | c6472714a2 |
14 | 65a488564b |
15 | 36a18c4801 |
hex | 1ed5b00ef3 |
132434104051 has 2 divisors, whose sum is σ = 132434104052. Its totient is φ = 132434104050.
The previous prime is 132434104027. The next prime is 132434104079. The reversal of 132434104051 is 150401434231.
It is an a-pointer prime, because the next prime (132434104079) can be obtained adding 132434104051 to its sum of digits (28).
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 132434104051 - 219 = 132433579763 is a prime.
It is a super-2 number, since 2×1324341040512 (a number of 23 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (132434104081) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 66217052025 + 66217052026.
It is an arithmetic number, because the mean of its divisors is an integer number (66217052026).
Almost surely, 2132434104051 is an apocalyptic number.
132434104051 is a deficient number, since it is larger than the sum of its proper divisors (1).
132434104051 is an equidigital number, since it uses as much as digits as its factorization.
132434104051 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 5760, while the sum is 28.
Adding to 132434104051 its reverse (150401434231), we get a palindrome (282835538282).
The spelling of 132434104051 in words is "one hundred thirty-two billion, four hundred thirty-four million, one hundred four thousand, fifty-one".
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