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51311052013493 is a prime number
BaseRepresentation
bin10111010101010110010010…
…00001110011111110110101
320201200021210102211011201002
423222223021001303332311
523211140013303412433
6301043545532013045
713544046462600266
oct1352531101637665
9221607712734632
1051311052013493
1115392993204628
125908513601185
1322827c445676c
14c9567a92296d
155deab8e9a9e8
hex2eaac9073fb5

51311052013493 has 2 divisors, whose sum is σ = 51311052013494. Its totient is φ = 51311052013492.

The previous prime is 51311052013483. The next prime is 51311052013517. The reversal of 51311052013493 is 39431025011315.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 48760026534244 + 2551025479249 = 6982838^2 + 1597193^2 .

It is a cyclic number.

It is not a de Polignac number, because 51311052013493 - 220 = 51311050964917 is a prime.

It is a super-2 number, since 2×513110520134932 (a number of 28 digits) contains 22 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (51311052013453) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 25655526006746 + 25655526006747.

It is an arithmetic number, because the mean of its divisors is an integer number (25655526006747).

Almost surely, 251311052013493 is an apocalyptic number.

It is an amenable number.

51311052013493 is a deficient number, since it is larger than the sum of its proper divisors (1).

51311052013493 is an equidigital number, since it uses as much as digits as its factorization.

51311052013493 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 48600, while the sum is 38.

The spelling of 51311052013493 in words is "fifty-one trillion, three hundred eleven billion, fifty-two million, thirteen thousand, four hundred ninety-three".