Base | Representation |
---|---|
bin | 1010001110011000101010… |
… | …1011110000011011100111 |
3 | 1110210202020011022200120022 |
4 | 2203212022223300123213 |
5 | 2433143120032440012 |
6 | 35524342101225355 |
7 | 2240140405326644 |
oct | 243461253603347 |
9 | 43722204280508 |
10 | 11242256140007 |
11 | 36448a5562572 |
12 | 13169a8a0225b |
13 | 6371a9052169 |
14 | 2ac1b1716dcb |
15 | 147684721472 |
hex | a398aaf06e7 |
11242256140007 has 2 divisors, whose sum is σ = 11242256140008. Its totient is φ = 11242256140006.
The previous prime is 11242256139979. The next prime is 11242256140009. The reversal of 11242256140007 is 70004165224211.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-11242256140007 is a prime.
Together with 11242256140009, it forms a pair of twin primes.
It is a Chen prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (11242256140009) 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, 5621128070003 + 5621128070004.
It is an arithmetic number, because the mean of its divisors is an integer number (5621128070004).
Almost surely, 211242256140007 is an apocalyptic number.
11242256140007 is a deficient number, since it is larger than the sum of its proper divisors (1).
11242256140007 is an equidigital number, since it uses as much as digits as its factorization.
11242256140007 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 26880, while the sum is 35.
The spelling of 11242256140007 in words is "eleven trillion, two hundred forty-two billion, two hundred fifty-six million, one hundred forty thousand, seven".
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