Base | Representation |
---|---|
bin | 10110011010111010110… |
… | …011111000100110110011 |
3 | 12110021220021221021001212 |
4 | 112122322303320212303 |
5 | 200220410002032342 |
6 | 3135445215350335 |
7 | 216212533533461 |
oct | 26327263704663 |
9 | 5407807837055 |
10 | 1540732455347 |
11 | 54446a338a28 |
12 | 20a7302223ab |
13 | b23a115c9a7 |
14 | 5480121d831 |
15 | 2a128315182 |
hex | 166bacf89b3 |
1540732455347 has 2 divisors, whose sum is σ = 1540732455348. Its totient is φ = 1540732455346.
The previous prime is 1540732455323. The next prime is 1540732455349. The reversal of 1540732455347 is 7435542370451.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-1540732455347 is a prime.
Together with 1540732455349, it forms a pair of twin primes.
It is a Chen prime.
It is a junction number, because it is equal to n+sod(n) for n = 1540732455295 and 1540732455304.
It is not a weakly prime, because it can be changed into another prime (1540732455349) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 770366227673 + 770366227674.
It is an arithmetic number, because the mean of its divisors is an integer number (770366227674).
Almost surely, 21540732455347 is an apocalyptic number.
1540732455347 is a deficient number, since it is larger than the sum of its proper divisors (1).
1540732455347 is an equidigital number, since it uses as much as digits as its factorization.
1540732455347 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 7056000, while the sum is 50.
The spelling of 1540732455347 in words is "one trillion, five hundred forty billion, seven hundred thirty-two million, four hundred fifty-five thousand, three hundred forty-seven".
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