Base | Representation |
---|---|
bin | 111111000101000111… |
… | …1110111111110001111 |
3 | 110221122200021220011011 |
4 | 1332022033313332033 |
5 | 4204412112402112 |
6 | 142121522510051 |
7 | 12533623634113 |
oct | 1761217677617 |
9 | 427580256134 |
10 | 135463403407 |
11 | 524a4617a62 |
12 | 22306505327 |
13 | ca0aa43b07 |
14 | 67b0d17743 |
15 | 37cc804aa7 |
hex | 1f8a3f7f8f |
135463403407 has 2 divisors, whose sum is σ = 135463403408. Its totient is φ = 135463403406.
The previous prime is 135463403377. The next prime is 135463403423. The reversal of 135463403407 is 704304364531.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-135463403407 is a prime.
It is a self number, because there is not a number n which added to its sum of digits gives 135463403407.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (135463403507) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 67731701703 + 67731701704.
It is an arithmetic number, because the mean of its divisors is an integer number (67731701704).
Almost surely, 2135463403407 is an apocalyptic number.
135463403407 is a deficient number, since it is larger than the sum of its proper divisors (1).
135463403407 is an equidigital number, since it uses as much as digits as its factorization.
135463403407 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 362880, while the sum is 40.
Adding to 135463403407 its reverse (704304364531), we get a palindrome (839767767938).
The spelling of 135463403407 in words is "one hundred thirty-five billion, four hundred sixty-three million, four hundred three thousand, four hundred seven".
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