Base | Representation |
---|---|
bin | 1001010100101111110001… |
… | …01100111110010111001011 |
3 | 2200121011112121011212002011 |
4 | 10222113320230332113023 |
5 | 10141414244132140042 |
6 | 111335233045520351 |
7 | 4214240336422666 |
oct | 452277054762713 |
9 | 80534477155064 |
10 | 20504051443147 |
11 | 65957a0555758 |
12 | 23719a54130b7 |
13 | b596a68c543c |
14 | 50c5895866dd |
15 | 258556501a17 |
hex | 12a5f8b3e5cb |
20504051443147 has 2 divisors, whose sum is σ = 20504051443148. Its totient is φ = 20504051443146.
The previous prime is 20504051443093. The next prime is 20504051443189. The reversal of 20504051443147 is 74134415040502.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 20504051443147 - 235 = 20469691704779 is a prime.
It is a super-2 number, since 2×205040514431472 (a number of 27 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (20504051443247) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 10252025721573 + 10252025721574.
It is an arithmetic number, because the mean of its divisors is an integer number (10252025721574).
Almost surely, 220504051443147 is an apocalyptic number.
20504051443147 is a deficient number, since it is larger than the sum of its proper divisors (1).
20504051443147 is an equidigital number, since it uses as much as digits as its factorization.
20504051443147 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 268800, while the sum is 40.
Adding to 20504051443147 its reverse (74134415040502), we get a palindrome (94638466483649).
The spelling of 20504051443147 in words is "twenty trillion, five hundred four billion, fifty-one million, four hundred forty-three thousand, one hundred forty-seven".
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