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51000113 is a prime number
BaseRepresentation
bin1100001010001…
…1001100110001
310112222002000002
43002203030301
5101024000423
65021035345
71156331243
oct302431461
9115862002
1051000113
112687415a
12150b5b55
13a7486a8
146ab8093
154726228
hex30a3331

51000113 has 2 divisors, whose sum is σ = 51000114. Its totient is φ = 51000112.

The previous prime is 51000083. The next prime is 51000127. The reversal of 51000113 is 31100015.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 50865424 + 134689 = 7132^2 + 367^2 .

It is a cyclic number.

It is not a de Polignac number, because 51000113 - 26 = 51000049 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 51000094 and 51000103.

It is not a weakly prime, because it can be changed into another prime (51000133) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (11) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 25500056 + 25500057.

It is an arithmetic number, because the mean of its divisors is an integer number (25500057).

Almost surely, 251000113 is an apocalyptic number.

It is an amenable number.

51000113 is a deficient number, since it is larger than the sum of its proper divisors (1).

51000113 is an equidigital number, since it uses as much as digits as its factorization.

51000113 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 15, while the sum is 11.

The square root of 51000113 is about 7141.4363401209. The cubic root of 51000113 is about 370.8432508170.

Adding to 51000113 its reverse (31100015), we get a palindrome (82100128).

The spelling of 51000113 in words is "fifty-one million, one hundred thirteen", and thus it is an aban number.