Base | Representation |
---|---|
bin | 10101011100111110010011… |
… | …11001001011000101100001 |
3 | 20012000212221202202120110111 |
4 | 22232133021321023011201 |
5 | 22140403411313120001 |
6 | 244155522442131321 |
7 | 12636165541324606 |
oct | 1256371171130541 |
9 | 205025852676414 |
10 | 47175013020001 |
11 | 140388a0aaa719 |
12 | 535aa004b0b41 |
13 | 204277a5b7099 |
14 | b913d60786ad |
15 | 56c1e4de0851 |
hex | 2ae7c9e4b161 |
47175013020001 has 2 divisors, whose sum is σ = 47175013020002. Its totient is φ = 47175013020000.
The previous prime is 47175013019971. The next prime is 47175013020073. The reversal of 47175013020001 is 10002031057174.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 36543657168225 + 10631355851776 = 6045135^2 + 3260576^2 .
It is a cyclic number.
It is not a de Polignac number, because 47175013020001 - 213 = 47175013011809 is a prime.
It is not a weakly prime, because it can be changed into another prime (47175013080001) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 23587506510000 + 23587506510001.
It is an arithmetic number, because the mean of its divisors is an integer number (23587506510001).
Almost surely, 247175013020001 is an apocalyptic number.
It is an amenable number.
47175013020001 is a deficient number, since it is larger than the sum of its proper divisors (1).
47175013020001 is an equidigital number, since it uses as much as digits as its factorization.
47175013020001 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 5880, while the sum is 31.
Adding to 47175013020001 its reverse (10002031057174), we get a palindrome (57177044077175).
The spelling of 47175013020001 in words is "forty-seven trillion, one hundred seventy-five billion, thirteen million, twenty thousand, one".
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