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5162241640613 is a prime number
BaseRepresentation
bin100101100011110110110…
…0101000100100010100101
3200021111122111101102020112
41023013231211010202211
51134034232324444423
614551300121542405
71041650105644541
oct113075545044245
920244574342215
105162241640613
111710326910472
126b4588344a05
132b5a49881a45
1413bbd4c96a21
158e435da3a78
hex4b1ed9448a5

5162241640613 has 2 divisors, whose sum is σ = 5162241640614. Its totient is φ = 5162241640612.

The previous prime is 5162241640579. The next prime is 5162241640631. The reversal of 5162241640613 is 3160461422615.

Together with next prime (5162241640631) it forms an Ormiston pair, because they use the same digits, order apart.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 5042979852964 + 119261787649 = 2245658^2 + 345343^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-5162241640613 is a prime.

It is a super-3 number, since 3×51622416406133 (a number of 39 digits) contains 333 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (5162241643613) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2581120820306 + 2581120820307.

It is an arithmetic number, because the mean of its divisors is an integer number (2581120820307).

Almost surely, 25162241640613 is an apocalyptic number.

It is an amenable number.

5162241640613 is a deficient number, since it is larger than the sum of its proper divisors (1).

5162241640613 is an equidigital number, since it uses as much as digits as its factorization.

5162241640613 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 207360, while the sum is 41.

The spelling of 5162241640613 in words is "five trillion, one hundred sixty-two billion, two hundred forty-one million, six hundred forty thousand, six hundred thirteen".