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1345110435287 is a prime number
BaseRepresentation
bin10011100100101110110…
…101000110010111010111
311202120222001121000000122
4103210232311012113113
5134014241232412122
62505534000502155
7166116043023413
oct23445665062727
94676861530018
101345110435287
11479504856973
1219883691a95b
1399ac6b63ab8
1449164763d43
1524ec93ca142
hex1392ed465d7

1345110435287 has 2 divisors, whose sum is σ = 1345110435288. Its totient is φ = 1345110435286.

The previous prime is 1345110435283. The next prime is 1345110435337. The reversal of 1345110435287 is 7825340115431.

It is a weak prime.

It is an emirp because it is prime and its reverse (7825340115431) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 1345110435287 - 22 = 1345110435283 is a prime.

It is a super-2 number, since 2×13451104352872 (a number of 25 digits) contains 22 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (1345110435283) 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, 672555217643 + 672555217644.

It is an arithmetic number, because the mean of its divisors is an integer number (672555217644).

Almost surely, 21345110435287 is an apocalyptic number.

1345110435287 is a deficient number, since it is larger than the sum of its proper divisors (1).

1345110435287 is an equidigital number, since it uses as much as digits as its factorization.

1345110435287 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 403200, while the sum is 44.

The spelling of 1345110435287 in words is "one trillion, three hundred forty-five billion, one hundred ten million, four hundred thirty-five thousand, two hundred eighty-seven".