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133414734799 is a prime number
BaseRepresentation
bin111110001000000100…
…0110100101111001111
3110202100220100202011001
41330100020310233033
54141213133003144
6141142340441131
712432066351253
oct1742010645717
9422326322131
10133414734799
1151643159a05
1221a343a91a7
13c772489378
146658bd0263
15370ca2bbd4
hex1f10234bcf

133414734799 has 2 divisors, whose sum is σ = 133414734800. Its totient is φ = 133414734798.

The previous prime is 133414734721. The next prime is 133414734823. The reversal of 133414734799 is 997437414331.

It is a strong prime.

It is an emirp because it is prime and its reverse (997437414331) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 133414734799 - 213 = 133414726607 is a prime.

It is a super-2 number, since 2×1334147347992 (a number of 23 digits) contains 22 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (133414734499) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 66707367399 + 66707367400.

It is an arithmetic number, because the mean of its divisors is an integer number (66707367400).

Almost surely, 2133414734799 is an apocalyptic number.

133414734799 is a deficient number, since it is larger than the sum of its proper divisors (1).

133414734799 is an equidigital number, since it uses as much as digits as its factorization.

133414734799 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 6858432, while the sum is 55.

The spelling of 133414734799 in words is "one hundred thirty-three billion, four hundred fourteen million, seven hundred thirty-four thousand, seven hundred ninety-nine".