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1354353171331 is a prime number
BaseRepresentation
bin10011101101010101101…
…111010101101110000011
311210110211012102122021211
4103231111233111232003
5134142203402440311
62514103052430551
7166564064003626
oct23552557255603
94713735378254
101354353171331
1148241806a2a6
1219a596178457
139a93a9b2394
14497a00c3dbd
152536aa53821
hex13b55bd5b83

1354353171331 has 2 divisors, whose sum is σ = 1354353171332. Its totient is φ = 1354353171330.

The previous prime is 1354353171317. The next prime is 1354353171343. The reversal of 1354353171331 is 1331713534531.

It is a strong prime.

It is an emirp because it is prime and its reverse (1331713534531) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 1354353171331 - 211 = 1354353169283 is a prime.

It is a super-2 number, since 2×13543531713312 (a number of 25 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (1354353172331) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 677176585665 + 677176585666.

It is an arithmetic number, because the mean of its divisors is an integer number (677176585666).

Almost surely, 21354353171331 is an apocalyptic number.

1354353171331 is a deficient number, since it is larger than the sum of its proper divisors (1).

1354353171331 is an equidigital number, since it uses as much as digits as its factorization.

1354353171331 is an evil number, because the sum of its binary digits is even.

The product of its digits is 170100, while the sum is 40.

The spelling of 1354353171331 in words is "one trillion, three hundred fifty-four billion, three hundred fifty-three million, one hundred seventy-one thousand, three hundred thirty-one".