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14325202304113 is a prime number
BaseRepresentation
bin1101000001110101100010…
…0110000011110001110001
31212201110211222102200101111
43100131120212003301301
53334201003242212423
650244523533033321
73005650603336261
oct320353046036161
955643758380344
1014325202304113
114623314274a98
1217343a1325841
137cbb2619c02c
143774b47797a1
1519c970de160d
hexd0758983c71

14325202304113 has 2 divisors, whose sum is σ = 14325202304114. Its totient is φ = 14325202304112.

The previous prime is 14325202304093. The next prime is 14325202304141. The reversal of 14325202304113 is 31140320252341.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 8486541969889 + 5838660334224 = 2913167^2 + 2416332^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-14325202304113 is a prime.

It is a super-2 number, since 2×143252023041132 (a number of 27 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (14325202364113) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7162601152056 + 7162601152057.

It is an arithmetic number, because the mean of its divisors is an integer number (7162601152057).

Almost surely, 214325202304113 is an apocalyptic number.

It is an amenable number.

14325202304113 is a deficient number, since it is larger than the sum of its proper divisors (1).

14325202304113 is an equidigital number, since it uses as much as digits as its factorization.

14325202304113 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 17280, while the sum is 31.

Adding to 14325202304113 its reverse (31140320252341), we get a palindrome (45465522556454).

The spelling of 14325202304113 in words is "fourteen trillion, three hundred twenty-five billion, two hundred two million, three hundred four thousand, one hundred thirteen".