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155114100131 is a prime number
BaseRepresentation
bin1001000001110110000…
…1010101010110100011
3112211101011112201101112
42100131201111112203
510020133202201011
6155131453253535
714130605045114
oct2203541252643
9484334481345
10155114100131
115a868958291
122608b4942ab
131181cc4a703
147716a7420b
15407ca7358b
hex241d8555a3

155114100131 has 2 divisors, whose sum is σ = 155114100132. Its totient is φ = 155114100130.

The previous prime is 155114100091. The next prime is 155114100179. The reversal of 155114100131 is 131001411551.

It is a weak prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-155114100131 is a prime.

It is a super-2 number, since 2×1551141001312 (a number of 23 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 155114100097 and 155114100106.

It is not a weakly prime, because it can be changed into another prime (155114106131) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 77557050065 + 77557050066.

It is an arithmetic number, because the mean of its divisors is an integer number (77557050066).

Almost surely, 2155114100131 is an apocalyptic number.

155114100131 is a deficient number, since it is larger than the sum of its proper divisors (1).

155114100131 is an equidigital number, since it uses as much as digits as its factorization.

155114100131 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 300, while the sum is 23.

Adding to 155114100131 its reverse (131001411551), we get a palindrome (286115511682).

The spelling of 155114100131 in words is "one hundred fifty-five billion, one hundred fourteen million, one hundred thousand, one hundred thirty-one".