Base | Representation |
---|---|
bin | 11000001010111010000001… |
… | …00110001010001101111111 |
3 | 20222012012210202122210122122 |
4 | 30011131000212022031333 |
5 | 23431312332313030101 |
6 | 305013212203233155 |
7 | 14124024601635332 |
oct | 1405350046121577 |
9 | 228165722583578 |
10 | 53151304033151 |
11 | 15a32383525365 |
12 | 5b650b42187bb |
13 | 23871b9a1cc7c |
14 | d1a773142619 |
15 | 6228c26bd51b |
hex | 30574098a37f |
53151304033151 has 2 divisors, whose sum is σ = 53151304033152. Its totient is φ = 53151304033150.
The previous prime is 53151304033129. The next prime is 53151304033211. The reversal of 53151304033151 is 15133040315135.
It is a weak prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-53151304033151 is a prime.
It is a super-2 number, since 2×531513040331512 (a number of 28 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (53151304033351) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 26575652016575 + 26575652016576.
It is an arithmetic number, because the mean of its divisors is an integer number (26575652016576).
Almost surely, 253151304033151 is an apocalyptic number.
53151304033151 is a deficient number, since it is larger than the sum of its proper divisors (1).
53151304033151 is an equidigital number, since it uses as much as digits as its factorization.
53151304033151 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 40500, while the sum is 35.
Adding to 53151304033151 its reverse (15133040315135), we get a palindrome (68284344348286).
The spelling of 53151304033151 in words is "fifty-three trillion, one hundred fifty-one billion, three hundred four million, thirty-three thousand, one hundred fifty-one".
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