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11312124352117 is a prime number
BaseRepresentation
bin1010010010011100111100…
…1001110101011001110101
31111001102120101200121102221
42210213033021311121311
52440314212313231432
640020415043402341
72245162662331231
oct244471711653165
944042511617387
1011312124352117
113671499321426
1213284476833b1
136409620a62c4
142b171ca255c1
151493c349db97
hexa49cf275675

11312124352117 has 2 divisors, whose sum is σ = 11312124352118. Its totient is φ = 11312124352116.

The previous prime is 11312124352097. The next prime is 11312124352141. The reversal of 11312124352117 is 71125342121311.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 8317144530916 + 2994979821201 = 2883946^2 + 1730601^2 .

It is a cyclic number.

It is not a de Polignac number, because 11312124352117 - 223 = 11312115963509 is a prime.

It is a super-3 number, since 3×113121243521173 (a number of 40 digits) contains 333 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (11312124352817) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5656062176058 + 5656062176059.

It is an arithmetic number, because the mean of its divisors is an integer number (5656062176059).

Almost surely, 211312124352117 is an apocalyptic number.

It is an amenable number.

11312124352117 is a deficient number, since it is larger than the sum of its proper divisors (1).

11312124352117 is an equidigital number, since it uses as much as digits as its factorization.

11312124352117 is an evil number, because the sum of its binary digits is even.

The product of its digits is 10080, while the sum is 34.

Adding to 11312124352117 its reverse (71125342121311), we get a palindrome (82437466473428).

The spelling of 11312124352117 in words is "eleven trillion, three hundred twelve billion, one hundred twenty-four million, three hundred fifty-two thousand, one hundred seventeen".