Search a number
-
+
51013319999977 is a prime number
BaseRepresentation
bin10111001100101011101101…
…10100010100000111101001
320200121211021222111210112101
423212111312310110013221
523141300234404444402
6300255110133413401
713513406424333256
oct1346256664240751
9220554258453471
1051013319999977
111528869a253125
12587a88179a261
1322606c64716ba
14c850b4ab842d
155d6e90987b87
hex2e6576d141e9

51013319999977 has 2 divisors, whose sum is σ = 51013319999978. Its totient is φ = 51013319999976.

The previous prime is 51013319999933. The next prime is 51013320000001. The reversal of 51013319999977 is 77999991331015.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 25905493985536 + 25107826014441 = 5089744^2 + 5010771^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-51013319999977 is a prime.

It is not a weakly prime, because it can be changed into another prime (51013319999077) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 25506659999988 + 25506659999989.

It is an arithmetic number, because the mean of its divisors is an integer number (25506659999989).

Almost surely, 251013319999977 is an apocalyptic number.

It is an amenable number.

51013319999977 is a deficient number, since it is larger than the sum of its proper divisors (1).

51013319999977 is an equidigital number, since it uses as much as digits as its factorization.

51013319999977 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 130203045, while the sum is 73.

The spelling of 51013319999977 in words is "fifty-one trillion, thirteen billion, three hundred nineteen million, nine hundred ninety-nine thousand, nine hundred seventy-seven".