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1034331121277 is a prime number
BaseRepresentation
bin11110000110100101111…
…00000110011001111101
310122212210022111002022022
433003102330012121331
5113421302231340102
62111055401325525
7134504455316426
oct17032274063175
93585708432268
101034331121277
11369725a534a3
121485634968a5
13766c9a5859c
14380c1b9a64d
151bd8a7205a2
hexf0d2f0667d

1034331121277 has 2 divisors, whose sum is σ = 1034331121278. Its totient is φ = 1034331121276.

The previous prime is 1034331121139. The next prime is 1034331121279. The reversal of 1034331121277 is 7721211334301.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 1004336691556 + 29994429721 = 1002166^2 + 173189^2 .

It is a cyclic number.

It is not a de Polignac number, because 1034331121277 - 232 = 1030036153981 is a prime.

It is a super-2 number, since 2×10343311212772 (a number of 25 digits) contains 22 as substring.

Together with 1034331121279, it forms a pair of twin primes.

It is a Chen prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (1034331121279) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 517165560638 + 517165560639.

It is an arithmetic number, because the mean of its divisors is an integer number (517165560639).

Almost surely, 21034331121277 is an apocalyptic number.

It is an amenable number.

1034331121277 is a deficient number, since it is larger than the sum of its proper divisors (1).

1034331121277 is an equidigital number, since it uses as much as digits as its factorization.

1034331121277 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 21168, while the sum is 35.

Adding to 1034331121277 its reverse (7721211334301), we get a palindrome (8755542455578).

The spelling of 1034331121277 in words is "one trillion, thirty-four billion, three hundred thirty-one million, one hundred twenty-one thousand, two hundred seventy-seven".