1340132243 has 4 divisors (see below), whose sum is σ = 1340236800. Its totient is φ = 1340027688.

The previous prime is 1340132239. The next prime is 1340132249. The reversal of 1340132243 is 3422310431.

It is a semiprime because it is the product of two primes, and also a brilliant number, because the two primes have the same length, and also an emirpimes, since its reverse is a distinct semiprime: 3422310431 = 257 ⋅13316383.

It is a cyclic number.

It is not a de Polignac number, because 1340132243 - 2² = 1340132239 is a prime.

It is a super-3 number, since 3×1340132243³ (a number of 28 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (1340132249) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 29843 + ... + 59756.

It is an arithmetic number, because the mean of its divisors is an integer number (335059200).

Almost surely, 2^1340132243 is an apocalyptic number.

1340132243 is a deficient number, since it is larger than the sum of its proper divisors (104557).

1340132243 is an equidigital number, since it uses as much as digits as its factorization.

1340132243 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 104556.

The product of its (nonzero) digits is 1728, while the sum is 23.

The square root of 1340132243 is about 36607.8166926136. The cubic root of 1340132243 is about 1102.5100375301.

Adding to 1340132243 its reverse (3422310431), we get a palindrome (4762442674).

The spelling of 1340132243 in words is "one billion, three hundred forty million, one hundred thirty-two thousand, two hundred forty-three".