Base | Representation |
---|---|
bin | 1001001010100001111111… |
… | …01001111100110011010111 |
3 | 2122100121120000112102112221 |
4 | 10211100333221330303113 |
5 | 10120141423423430031 |
6 | 110510104015323211 |
7 | 4150003322644621 |
oct | 445207751746327 |
9 | 78317500472487 |
10 | 20153054514391 |
11 | 646a953980047 |
12 | 2315969220507 |
13 | b3256a623856 |
14 | 4d95ad565c11 |
15 | 24e361bc6a11 |
hex | 12543fa7ccd7 |
20153054514391 has 2 divisors, whose sum is σ = 20153054514392. Its totient is φ = 20153054514390.
The previous prime is 20153054514349. The next prime is 20153054514421. The reversal of 20153054514391 is 19341545035102.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-20153054514391 is a prime.
It is a super-2 number, since 2×201530545143912 (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 (20153054514991) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 10076527257195 + 10076527257196.
It is an arithmetic number, because the mean of its divisors is an integer number (10076527257196).
Almost surely, 220153054514391 is an apocalyptic number.
20153054514391 is a deficient number, since it is larger than the sum of its proper divisors (1).
20153054514391 is an equidigital number, since it uses as much as digits as its factorization.
20153054514391 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 324000, while the sum is 43.
Adding to 20153054514391 its reverse (19341545035102), we get a palindrome (39494599549493).
The spelling of 20153054514391 in words is "twenty trillion, one hundred fifty-three billion, fifty-four million, five hundred fourteen thousand, three hundred ninety-one".
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