Base | Representation |
---|---|
bin | 11110110010101011101000… |
… | …110010001100000001101011 |
3 | 122202111102121001000102002022 |
4 | 132302223220302030001223 |
5 | 120222242302411012301 |
6 | 1200005010233455055 |
7 | 40345035344505323 |
oct | 3662535062140153 |
9 | 582442531012068 |
10 | 135424224313451 |
11 | 3a172100139347 |
12 | 1323215687148b |
13 | 5a745ac8a1968 |
14 | 25627d3054283 |
15 | 109ca65212d1b |
hex | 7b2ae8c8c06b |
135424224313451 has 2 divisors, whose sum is σ = 135424224313452. Its totient is φ = 135424224313450.
The previous prime is 135424224313373. The next prime is 135424224313453. The reversal of 135424224313451 is 154313422424531.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-135424224313451 is a prime.
Together with 135424224313453, it forms a pair of twin primes.
It is a Chen prime.
It is not a weakly prime, because it can be changed into another prime (135424224313453) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 67712112156725 + 67712112156726.
It is an arithmetic number, because the mean of its divisors is an integer number (67712112156726).
Almost surely, 2135424224313451 is an apocalyptic number.
135424224313451 is a deficient number, since it is larger than the sum of its proper divisors (1).
135424224313451 is an equidigital number, since it uses as much as digits as its factorization.
135424224313451 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 1382400, while the sum is 44.
Adding to 135424224313451 its reverse (154313422424531), we get a palindrome (289737646737982).
The spelling of 135424224313451 in words is "one hundred thirty-five trillion, four hundred twenty-four billion, two hundred twenty-four million, three hundred thirteen thousand, four hundred fifty-one".
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