Base | Representation |
---|---|
bin | 100101011000000100… |
… | …110001010111011011 |
3 | 10211120220002200120002 |
4 | 211120010301113123 |
5 | 1124142340221301 |
6 | 30234143323215 |
7 | 2620342234043 |
oct | 453004612733 |
9 | 124526080502 |
10 | 40132351451 |
11 | 16024736705 |
12 | 79402b050b |
13 | 3a2761242a |
14 | 1d29dc0723 |
15 | 109d42c16b |
hex | 9581315db |
40132351451 has 2 divisors, whose sum is σ = 40132351452. Its totient is φ = 40132351450.
The previous prime is 40132351351. The next prime is 40132351453. The reversal of 40132351451 is 15415323104.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 40132351451 - 230 = 39058609627 is a prime.
It is a super-2 number, since 2×401323514512 (a number of 22 digits) contains 22 as substring.
Together with 40132351453, it forms a pair of twin primes.
It is a Chen prime.
It is not a weakly prime, because it can be changed into another prime (40132351453) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (17) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 20066175725 + 20066175726.
It is an arithmetic number, because the mean of its divisors is an integer number (20066175726).
Almost surely, 240132351451 is an apocalyptic number.
40132351451 is a deficient number, since it is larger than the sum of its proper divisors (1).
40132351451 is an equidigital number, since it uses as much as digits as its factorization.
40132351451 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 7200, while the sum is 29.
Adding to 40132351451 its reverse (15415323104), we get a palindrome (55547674555).
The spelling of 40132351451 in words is "forty billion, one hundred thirty-two million, three hundred fifty-one thousand, four hundred fifty-one".
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