Base | Representation |
---|---|
bin | 11110011110100100100100… |
… | …001110111100101110101111 |
3 | 122120121022110022221002102102 |
4 | 132132210210032330232233 |
5 | 120032122003003130112 |
6 | 1153030053405445315 |
7 | 40143142501630535 |
oct | 3636444416745657 |
9 | 576538408832372 |
10 | 134042242239407 |
11 | 39789aa52a7557 |
12 | 1304a34b87083b |
13 | 59a418ab33748 |
14 | 25159717c0155 |
15 | 1076b2dcecbc2 |
hex | 79e9243bcbaf |
134042242239407 has 2 divisors, whose sum is σ = 134042242239408. Its totient is φ = 134042242239406.
The previous prime is 134042242239403. The next prime is 134042242239443. The reversal of 134042242239407 is 704932242240431.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 134042242239407 - 22 = 134042242239403 is a prime.
It is a super-2 number, since 2×1340422422394072 (a number of 29 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (134042242239403) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 67021121119703 + 67021121119704.
It is an arithmetic number, because the mean of its divisors is an integer number (67021121119704).
Almost surely, 2134042242239407 is an apocalyptic number.
134042242239407 is a deficient number, since it is larger than the sum of its proper divisors (1).
134042242239407 is an equidigital number, since it uses as much as digits as its factorization.
134042242239407 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 2322432, while the sum is 47.
Adding to 134042242239407 its reverse (704932242240431), we get a palindrome (838974484479838).
The spelling of 134042242239407 in words is "one hundred thirty-four trillion, forty-two billion, two hundred forty-two million, two hundred thirty-nine thousand, four hundred seven".
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