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14312433221953 is a prime number
BaseRepresentation
bin1101000001000101111101…
…1111111000000101000001
31212200020220002011002122121
43100101133133320011001
53333443330401100303
650235012511204241
73005016303046426
oct320213737700501
955606802132577
1014312433221953
11461896147a629
121731a18b16681
137ca86c8bb309
14376a2295dd4d
1519c474da49bd
hexd045f7f8141

14312433221953 has 2 divisors, whose sum is σ = 14312433221954. Its totient is φ = 14312433221952.

The previous prime is 14312433221941. The next prime is 14312433222179. The reversal of 14312433221953 is 35912233421341.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 13874939608464 + 437493613489 = 3724908^2 + 661433^2 .

It is a cyclic number.

It is not a de Polignac number, because 14312433221953 - 217 = 14312433090881 is a prime.

It is a super-2 number, since 2×143124332219532 (a number of 27 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (14312433221903) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7156216610976 + 7156216610977.

It is an arithmetic number, because the mean of its divisors is an integer number (7156216610977).

Almost surely, 214312433221953 is an apocalyptic number.

It is an amenable number.

14312433221953 is a deficient number, since it is larger than the sum of its proper divisors (1).

14312433221953 is an equidigital number, since it uses as much as digits as its factorization.

14312433221953 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 466560, while the sum is 43.

The spelling of 14312433221953 in words is "fourteen trillion, three hundred twelve billion, four hundred thirty-three million, two hundred twenty-one thousand, nine hundred fifty-three".