Base | Representation |
---|---|
bin | 1011111101000011000001… |
… | …1001111001111101101111 |
3 | 1201112111102222112010212111 |
4 | 2333100300121321331233 |
5 | 3210320201434220101 |
6 | 43542000005544451 |
7 | 2524403000602144 |
oct | 277206031717557 |
9 | 51474388463774 |
10 | 13143412023151 |
11 | 42080a9491383 |
12 | 1583346115127 |
13 | 7445593c2907 |
14 | 3362048703cb |
15 | 17bd54bd8251 |
hex | bf430679f6f |
13143412023151 has 2 divisors, whose sum is σ = 13143412023152. Its totient is φ = 13143412023150.
The previous prime is 13143412023097. The next prime is 13143412023179. The reversal of 13143412023151 is 15132021434131.
It is a happy number.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 13143412023151 - 213 = 13143412014959 is a prime.
It is a super-2 number, since 2×131434120231512 (a number of 27 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (13143412023451) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6571706011575 + 6571706011576.
It is an arithmetic number, because the mean of its divisors is an integer number (6571706011576).
Almost surely, 213143412023151 is an apocalyptic number.
13143412023151 is a deficient number, since it is larger than the sum of its proper divisors (1).
13143412023151 is an equidigital number, since it uses as much as digits as its factorization.
13143412023151 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 8640, while the sum is 31.
Adding to 13143412023151 its reverse (15132021434131), we get a palindrome (28275433457282).
The spelling of 13143412023151 in words is "thirteen trillion, one hundred forty-three billion, four hundred twelve million, twenty-three thousand, one hundred fifty-one".
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