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14311055524321 is a prime number
BaseRepresentation
bin1101000001000000110101…
…1000011000000111100001
31212200010100001202210210101
43100100031120120013201
53333433020203234241
650234224054302401
73004640203602313
oct320201530300741
955603301683711
1014311055524321
114618314833a46
1217316b3657a01
137ca6b034ac0a
143769319b57b3
1519c3e3e67e31
hexd040d6181e1

14311055524321 has 2 divisors, whose sum is σ = 14311055524322. Its totient is φ = 14311055524320.

The previous prime is 14311055524313. The next prime is 14311055524351. The reversal of 14311055524321 is 12342555011341.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 11631073710096 + 2679981814225 = 3410436^2 + 1637065^2 .

It is a cyclic number.

It is not a de Polignac number, because 14311055524321 - 23 = 14311055524313 is a prime.

It is a super-2 number, since 2×143110555243212 (a number of 27 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (14311055524351) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7155527762160 + 7155527762161.

It is an arithmetic number, because the mean of its divisors is an integer number (7155527762161).

Almost surely, 214311055524321 is an apocalyptic number.

It is an amenable number.

14311055524321 is a deficient number, since it is larger than the sum of its proper divisors (1).

14311055524321 is an equidigital number, since it uses as much as digits as its factorization.

14311055524321 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 72000, while the sum is 37.

The spelling of 14311055524321 in words is "fourteen trillion, three hundred eleven billion, fifty-five million, five hundred twenty-four thousand, three hundred twenty-one".