1134331540142 has 4 divisors (see below), whose sum is σ = 1701497310216. Its totient is φ = 567165770070.

The previous prime is 1134331540127. The next prime is 1134331540157. The reversal of 1134331540142 is 2410451334311.

It is a semiprime because it is the product of two primes.

It is an interprime number because it is at equal distance from previous prime (1134331540127) and next prime (1134331540157).

It is a junction number, because it is equal to n+sod(n) for n = 1134331540099 and 1134331540108.

It is a congruent number.

It is an unprimeable number.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 283582885034 + ... + 283582885037.

It is an arithmetic number, because the mean of its divisors is an integer number (425374327554).

Almost surely, 2^{1134331540142} is an apocalyptic number.

1134331540142 is a deficient number, since it is larger than the sum of its proper divisors (567165770074).

1134331540142 is an equidigital number, since it uses as much as digits as its factorization.

1134331540142 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 567165770073.

The product of its (nonzero) digits is 17280, while the sum is 32.

Adding to 1134331540142 its reverse (2410451334311), we get a palindrome (3544782874453).

The spelling of 1134331540142 in words is "one trillion, one hundred thirty-four billion, three hundred thirty-one million, five hundred forty thousand, one hundred forty-two".