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314543005214351 is a prime number
BaseRepresentation
bin100011110000100110100000…
…0000101111010011010001111
31112020201000110211222111102212
41013201031000011322122033
5312211433033313324401
63032551021201341035
7123152660615251634
oct10741150005723217
91466630424874385
10314543005214351
119124aa37362671
122b34066003777b
1310668380047b14
145795d6985818b
152656ebaaeadbb
hex11e134017a68f

314543005214351 has 2 divisors, whose sum is σ = 314543005214352. Its totient is φ = 314543005214350.

The previous prime is 314543005214329. The next prime is 314543005214353. The reversal of 314543005214351 is 153412500345413.

It is a strong prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-314543005214351 is a prime.

Together with 314543005214353, it forms a pair of twin primes.

It is a Chen prime.

It is a junction number, because it is equal to n+sod(n) for n = 314543005214299 and 314543005214308.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (314543005214353) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 157271502607175 + 157271502607176.

It is an arithmetic number, because the mean of its divisors is an integer number (157271502607176).

Almost surely, 2314543005214351 is an apocalyptic number.

314543005214351 is a deficient number, since it is larger than the sum of its proper divisors (1).

314543005214351 is an equidigital number, since it uses as much as digits as its factorization.

314543005214351 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 432000, while the sum is 41.

Adding to 314543005214351 its reverse (153412500345413), we get a palindrome (467955505559764).

The spelling of 314543005214351 in words is "three hundred fourteen trillion, five hundred forty-three billion, five million, two hundred fourteen thousand, three hundred fifty-one".