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131221314430307 is a prime number
BaseRepresentation
bin11101110101100001010111…
…110101010110010101100011
3122012121122011000120122101222
4131311201113311112111203
5114144412222443232212
61143030123432445255
736432266333112653
oct3565412765262543
9565548130518358
10131221314430307
11388a1713077282
12128736a084982b
13582b169965a1a
1424592073a8363
15102857a5e0c72
hex775857d56563

131221314430307 has 2 divisors, whose sum is σ = 131221314430308. Its totient is φ = 131221314430306.

The previous prime is 131221314430207. The next prime is 131221314430321. The reversal of 131221314430307 is 703034413122131.

It is a happy number.

It is a strong prime.

It is an emirp because it is prime and its reverse (703034413122131) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 131221314430307 - 220 = 131221313381731 is a prime.

It is a super-3 number, since 3×1312213144303073 (a number of 43 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is not a weakly prime, because it can be changed into another prime (131221314430357) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65610657215153 + 65610657215154.

It is an arithmetic number, because the mean of its divisors is an integer number (65610657215154).

Almost surely, 2131221314430307 is an apocalyptic number.

131221314430307 is a deficient number, since it is larger than the sum of its proper divisors (1).

131221314430307 is an equidigital number, since it uses as much as digits as its factorization.

131221314430307 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 36288, while the sum is 35.

Adding to 131221314430307 its reverse (703034413122131), we get a palindrome (834255727552438).

The spelling of 131221314430307 in words is "one hundred thirty-one trillion, two hundred twenty-one billion, three hundred fourteen million, four hundred thirty thousand, three hundred seven".