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47175013020001 is a prime number
BaseRepresentation
bin10101011100111110010011…
…11001001011000101100001
320012000212221202202120110111
422232133021321023011201
522140403411313120001
6244155522442131321
712636165541324606
oct1256371171130541
9205025852676414
1047175013020001
11140388a0aaa719
12535aa004b0b41
13204277a5b7099
14b913d60786ad
1556c1e4de0851
hex2ae7c9e4b161

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".