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4334214555131 is a prime number
BaseRepresentation
bin111111000100100011010…
…100011010010111111011
3120100100100212100012210212
4333010203110122113323
51032002432411231011
613115035523501335
7625064606566166
oct77044324322773
916310325305725
104334214555131
11142114813a60a
1259bbbb82784b
1325593a1c3126
1410dac4530add
1577b2229a08b
hex3f12351a5fb

4334214555131 has 2 divisors, whose sum is σ = 4334214555132. Its totient is φ = 4334214555130.

The previous prime is 4334214555113. The next prime is 4334214555149. The reversal of 4334214555131 is 1315554124334.

Together with previous prime (4334214555113) it forms an Ormiston pair, because they use the same digits, order apart.

It is a balanced prime because it is at equal distance from previous prime (4334214555113) and next prime (4334214555149).

It is a cyclic number.

It is not a de Polignac number, because 4334214555131 - 26 = 4334214555067 is a prime.

It is a super-2 number, since 2×43342145551312 (a number of 26 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (4334214552131) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2167107277565 + 2167107277566.

It is an arithmetic number, because the mean of its divisors is an integer number (2167107277566).

Almost surely, 24334214555131 is an apocalyptic number.

4334214555131 is a deficient number, since it is larger than the sum of its proper divisors (1).

4334214555131 is an equidigital number, since it uses as much as digits as its factorization.

4334214555131 is an evil number, because the sum of its binary digits is even.

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

Adding to 4334214555131 its reverse (1315554124334), we get a palindrome (5649768679465).

The spelling of 4334214555131 in words is "four trillion, three hundred thirty-four billion, two hundred fourteen million, five hundred fifty-five thousand, one hundred thirty-one".