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132014004424117 is a prime number
BaseRepresentation
bin11110000001000011100111…
…110101101100110110110101
3122022102102012221122121001121
4132001003213311230312311
5114300404140113032432
61144434221342223541
736543461044642414
oct3601034765546665
9568372187577047
10132014004424117
1139077908999142
1212981246b84bb1
135887b3731703a
14248572495747b
15103dec1c80197
hex7810e7d6cdb5

132014004424117 has 2 divisors, whose sum is σ = 132014004424118. Its totient is φ = 132014004424116.

The previous prime is 132014004424073. The next prime is 132014004424121. The reversal of 132014004424117 is 711424400410231.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 117706312356516 + 14307692067601 = 10849254^2 + 3782551^2 .

It is a cyclic number.

It is not a de Polignac number, because 132014004424117 - 219 = 132014003899829 is a prime.

It is a super-2 number, since 2×1320140044241172 (a number of 29 digits) contains 22 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (132014004224117) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 66007002212058 + 66007002212059.

It is an arithmetic number, because the mean of its divisors is an integer number (66007002212059).

Almost surely, 2132014004424117 is an apocalyptic number.

It is an amenable number.

132014004424117 is a deficient number, since it is larger than the sum of its proper divisors (1).

132014004424117 is an equidigital number, since it uses as much as digits as its factorization.

132014004424117 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 21504, while the sum is 34.

Adding to 132014004424117 its reverse (711424400410231), we get a palindrome (843438404834348).

The spelling of 132014004424117 in words is "one hundred thirty-two trillion, fourteen billion, four million, four hundred twenty-four thousand, one hundred seventeen".