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11315305551343 is a prime number
BaseRepresentation
bin1010010010101000110011…
…0001001001010111101111
31111001201210002200021101111
42210222030301021113233
52440342221210120333
640022102443452451
72245334551145443
oct244521461112757
944051702607344
1011315305551343
113672880aa34a2
121328b94b27127
1364104b191501
142b1941318823
1514950c8d64cd
hexa4a8cc495ef

11315305551343 has 2 divisors, whose sum is σ = 11315305551344. Its totient is φ = 11315305551342.

The previous prime is 11315305551299. The next prime is 11315305551353. The reversal of 11315305551343 is 34315550351311.

11315305551343 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a strong prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-11315305551343 is a prime.

It is a super-2 number, since 2×113153055513432 (a number of 27 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 11315305551296 and 11315305551305.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (11315305551353) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5657652775671 + 5657652775672.

It is an arithmetic number, because the mean of its divisors is an integer number (5657652775672).

Almost surely, 211315305551343 is an apocalyptic number.

11315305551343 is a deficient number, since it is larger than the sum of its proper divisors (1).

11315305551343 is an equidigital number, since it uses as much as digits as its factorization.

11315305551343 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 202500, while the sum is 40.

The spelling of 11315305551343 in words is "eleven trillion, three hundred fifteen billion, three hundred five million, five hundred fifty-one thousand, three hundred forty-three".