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51330663363251 is a prime number
BaseRepresentation
bin10111010101111010110011…
…11101001011001010110011
320201202011102012000222001102
423222331121331023022303
523212000144310111001
6301100551535155015
713545342455403146
oct1352753175131263
9221664365028042
1051330663363251
11153a023730a9a8
125910287375a6b
1322845cb464b2c
14c965bb35a05d
155e0365a45e6b
hex2eaf59f4b2b3

51330663363251 has 2 divisors, whose sum is σ = 51330663363252. Its totient is φ = 51330663363250.

The previous prime is 51330663363229. The next prime is 51330663363253. The reversal of 51330663363251 is 15236336603315.

It is a strong prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-51330663363251 is a prime.

It is a super-2 number, since 2×513306633632512 (a number of 28 digits) contains 22 as substring.

Together with 51330663363253, it forms a pair of twin primes.

It is a Chen prime.

It is a junction number, because it is equal to n+sod(n) for n = 51330663363196 and 51330663363205.

It is not a weakly prime, because it can be changed into another prime (51330663363253) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 25665331681625 + 25665331681626.

It is an arithmetic number, because the mean of its divisors is an integer number (25665331681626).

Almost surely, 251330663363251 is an apocalyptic number.

51330663363251 is a deficient number, since it is larger than the sum of its proper divisors (1).

51330663363251 is an equidigital number, since it uses as much as digits as its factorization.

51330663363251 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 2624400, while the sum is 47.

Adding to 51330663363251 its reverse (15236336603315), we get a palindrome (66566999966566).

The spelling of 51330663363251 in words is "fifty-one trillion, three hundred thirty billion, six hundred sixty-three million, three hundred sixty-three thousand, two hundred fifty-one".