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140130247201279 is a prime number
BaseRepresentation
bin11111110111001010011101…
…100110110101000111111111
3200101011022122212022220121221
4133313022131212311013333
5121331343221240420104
61214010540304334211
741342035105160626
oct3767123546650777
9611138585286557
10140130247201279
11407169a43a2355
1213872220314367
1360262c1887a7a
1426864a95446bd
151130198dcec54
hex7f729d9b51ff

140130247201279 has 2 divisors, whose sum is σ = 140130247201280. Its totient is φ = 140130247201278.

The previous prime is 140130247201237. The next prime is 140130247201363. The reversal of 140130247201279 is 972102742031041.

It is a weak prime.

It is a cyclic number.

It is not a de Polignac number, because 140130247201279 - 211 = 140130247199231 is a prime.

It is a super-3 number, since 3×1401302472012793 (a number of 43 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (140130247204279) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 70065123600639 + 70065123600640.

It is an arithmetic number, because the mean of its divisors is an integer number (70065123600640).

Almost surely, 2140130247201279 is an apocalyptic number.

140130247201279 is a deficient number, since it is larger than the sum of its proper divisors (1).

140130247201279 is an equidigital number, since it uses as much as digits as its factorization.

140130247201279 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 169344, while the sum is 43.

The spelling of 140130247201279 in words is "one hundred forty trillion, one hundred thirty billion, two hundred forty-seven million, two hundred one thousand, two hundred seventy-nine".