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1331415258097 is a prime number
BaseRepresentation
bin10011010111111110100…
…010001001001111110001
311201021121121200220201211
4103113332202021033301
5133303214301224342
62455351013400121
7165122465652211
oct23277642111761
94637547626654
101331415258097
11473717226611
12196054339041
1398723750c62
1448625958241
1524976e09c17
hex135fe8893f1

1331415258097 has 2 divisors, whose sum is σ = 1331415258098. Its totient is φ = 1331415258096.

The previous prime is 1331415258059. The next prime is 1331415258119. The reversal of 1331415258097 is 7908525141331.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 911091249121 + 420324008976 = 954511^2 + 648324^2 .

It is a cyclic number.

It is not a de Polignac number, because 1331415258097 - 211 = 1331415256049 is a prime.

It is a super-2 number, since 2×13314152580972 (a number of 25 digits) contains 22 as substring.

It is a self number, because there is not a number n which added to its sum of digits gives 1331415258097.

It is not a weakly prime, because it can be changed into another prime (1331415258497) 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, 665707629048 + 665707629049.

It is an arithmetic number, because the mean of its divisors is an integer number (665707629049).

Almost surely, 21331415258097 is an apocalyptic number.

It is an amenable number.

1331415258097 is a deficient number, since it is larger than the sum of its proper divisors (1).

1331415258097 is an equidigital number, since it uses as much as digits as its factorization.

1331415258097 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 907200, while the sum is 49.

The spelling of 1331415258097 in words is "one trillion, three hundred thirty-one billion, four hundred fifteen million, two hundred fifty-eight thousand, ninety-seven".