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33312114122117 is a prime number
BaseRepresentation
bin1111001001100000101001…
…01110101001110110000101
311100221121101111202020010212
413210300110232221312011
513331241202203401432
6154503212435325205
710005502146043145
oct744602456516605
9140847344666125
1033312114122117
11a683651a822a6
1238a013b750805
13157841b77bc3c
1483245a22c925
153cb7d04912b2
hex1e4c14ba9d85

33312114122117 has 2 divisors, whose sum is σ = 33312114122118. Its totient is φ = 33312114122116.

The previous prime is 33312114122083. The next prime is 33312114122141. The reversal of 33312114122117 is 71122141121333.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 33294677183716 + 17436938401 = 5770154^2 + 132049^2 .

It is a cyclic number.

It is not a de Polignac number, because 33312114122117 - 210 = 33312114121093 is a prime.

It is a super-2 number, since 2×333121141221172 (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 (33312114422117) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 16656057061058 + 16656057061059.

It is an arithmetic number, because the mean of its divisors is an integer number (16656057061059).

Almost surely, 233312114122117 is an apocalyptic number.

It is an amenable number.

33312114122117 is a deficient number, since it is larger than the sum of its proper divisors (1).

33312114122117 is an equidigital number, since it uses as much as digits as its factorization.

33312114122117 is an evil number, because the sum of its binary digits is even.

The product of its digits is 6048, while the sum is 32.

The spelling of 33312114122117 in words is "thirty-three trillion, three hundred twelve billion, one hundred fourteen million, one hundred twenty-two thousand, one hundred seventeen".