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115251203437 is a prime number
BaseRepresentation
bin110101101010110000…
…0011100110101101101
3102000111001110202021021
41223111200130311231
53342013302002222
6124540133134141
711220014031316
oct1532540346555
9360431422237
10115251203437
1144972332a67
121a40548a351
13ab393ccb86
14581479560d
152ee811eac7
hex1ad581cd6d

115251203437 has 2 divisors, whose sum is σ = 115251203438. Its totient is φ = 115251203436.

The previous prime is 115251203399. The next prime is 115251203461. The reversal of 115251203437 is 734302152511.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 59316115401 + 55935088036 = 243549^2 + 236506^2 .

It is an emirp because it is prime and its reverse (734302152511) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 115251203437 - 219 = 115250679149 is a prime.

It is a super-2 number, since 2×1152512034372 (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 (115257203437) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 57625601718 + 57625601719.

It is an arithmetic number, because the mean of its divisors is an integer number (57625601719).

Almost surely, 2115251203437 is an apocalyptic number.

It is an amenable number.

115251203437 is a deficient number, since it is larger than the sum of its proper divisors (1).

115251203437 is an equidigital number, since it uses as much as digits as its factorization.

115251203437 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 25200, while the sum is 34.

Adding to 115251203437 its reverse (734302152511), we get a palindrome (849553355948).

The spelling of 115251203437 in words is "one hundred fifteen billion, two hundred fifty-one million, two hundred three thousand, four hundred thirty-seven".