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32121344312117 is a prime number
BaseRepresentation
bin1110100110110110101010…
…10100010100001100110101
311012201202210201021100111022
413103123111110110030311
513202234003120441432
6152152201501550525
76523455566465054
oct723332524241465
9135652721240438
1032121344312117
11a26464950a605
1237293b9275445
1314c005053c128
147d097791a39b
153aa83aca1b12
hex1d36d5514335

32121344312117 has 2 divisors, whose sum is σ = 32121344312118. Its totient is φ = 32121344312116.

The previous prime is 32121344312111. The next prime is 32121344312149. The reversal of 32121344312117 is 71121344312123.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 21000946817041 + 11120397495076 = 4582679^2 + 3334726^2 .

It is a cyclic number.

It is not a de Polignac number, because 32121344312117 - 222 = 32121340117813 is a prime.

It is a super-2 number, since 2×321213443121172 (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 (32121344312111) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 16060672156058 + 16060672156059.

It is an arithmetic number, because the mean of its divisors is an integer number (16060672156059).

Almost surely, 232121344312117 is an apocalyptic number.

It is an amenable number.

32121344312117 is a deficient number, since it is larger than the sum of its proper divisors (1).

32121344312117 is an equidigital number, since it uses as much as digits as its factorization.

32121344312117 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 24192, while the sum is 35.

The spelling of 32121344312117 in words is "thirty-two trillion, one hundred twenty-one billion, three hundred forty-four million, three hundred twelve thousand, one hundred seventeen".