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14311223301103 is a prime number
BaseRepresentation
bin1101000001000001011101…
…1000011001001111101111
31212200010202202110202021021
43100100113120121033233
53333433341131113403
650234252450315011
73004644311641411
oct320202730311757
955603682422237
1014311223301103
1146183a0508143
12173173b888a67
137ca709032311
14376949daa9b1
1519c403a598bd
hexd04176193ef

14311223301103 has 2 divisors, whose sum is σ = 14311223301104. Its totient is φ = 14311223301102.

The previous prime is 14311223301101. The next prime is 14311223301121. The reversal of 14311223301103 is 30110332211341.

14311223301103 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a weak prime.

It is a cyclic number.

It is not a de Polignac number, because 14311223301103 - 21 = 14311223301101 is a prime.

It is a super-2 number, since 2×143112233011032 (a number of 27 digits) contains 22 as substring.

Together with 14311223301101, it forms a pair of twin primes.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (14311223301101) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7155611650551 + 7155611650552.

It is an arithmetic number, because the mean of its divisors is an integer number (7155611650552).

Almost surely, 214311223301103 is an apocalyptic number.

14311223301103 is a deficient number, since it is larger than the sum of its proper divisors (1).

14311223301103 is an equidigital number, since it uses as much as digits as its factorization.

14311223301103 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 1296, while the sum is 25.

Adding to 14311223301103 its reverse (30110332211341), we get a palindrome (44421555512444).

The spelling of 14311223301103 in words is "fourteen trillion, three hundred eleven billion, two hundred twenty-three million, three hundred one thousand, one hundred three".