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64138851477133 is a prime number
BaseRepresentation
bin11101001010101011111010…
…11101010110011010001101
322102002121120211122002110121
432211111331131112122031
531401322331434232013
6344224552005304541
716336610350141636
oct1645257535263215
9272077524562417
1064138851477133
1119489152620664
12723a657705151
1329a33632c9284
1411ba4a093488d
157635eaa1158d
hex3a557d75668d

64138851477133 has 2 divisors, whose sum is σ = 64138851477134. Its totient is φ = 64138851477132.

The previous prime is 64138851477119. The next prime is 64138851477167. The reversal of 64138851477133 is 33177415883146.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 39905019629764 + 24233831847369 = 6317042^2 + 4922787^2 .

It is a cyclic number.

It is not a de Polignac number, because 64138851477133 - 217 = 64138851346061 is a prime.

It is a super-2 number, since 2×641388514771332 (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 (64138851477433) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 32069425738566 + 32069425738567.

It is an arithmetic number, because the mean of its divisors is an integer number (32069425738567).

Almost surely, 264138851477133 is an apocalyptic number.

It is an amenable number.

64138851477133 is a deficient number, since it is larger than the sum of its proper divisors (1).

64138851477133 is an equidigital number, since it uses as much as digits as its factorization.

64138851477133 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 40642560, while the sum is 61.

The spelling of 64138851477133 in words is "sixty-four trillion, one hundred thirty-eight billion, eight hundred fifty-one million, four hundred seventy-seven thousand, one hundred thirty-three".