101132113 has 2 divisors, whose sum is σ = 101132114. Its totient is φ = 101132112.

The previous prime is 101132107. The next prime is 101132147. The reversal of 101132113 is 311231101.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 86638864 + 14493249 = 9308^2 + 3807^2 .

It is a cyclic number.

It is not a de Polignac number, because 101132113 - 2¹⁷ = 101001041 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 101132093 and 101132102.

It is not a weakly prime, because it can be changed into another prime (101132153) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 50566056 + 50566057.

It is an arithmetic number, because the mean of its divisors is an integer number (50566057).

Almost surely, 2^101132113 is an apocalyptic number.

It is an amenable number.

101132113 is a deficient number, since it is larger than the sum of its proper divisors (1).

101132113 is an equidigital number, since it uses as much as digits as its factorization.

101132113 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 18, while the sum is 13.

The square root of 101132113 is about 10056.4463405320. The cubic root of 101132113 is about 465.9039156095.

Adding to 101132113 its reverse (311231101), we get a palindrome (412363214).

It can be divided in two parts, 1011 and 32113, that added together give a square (33124 = 182²).

The spelling of 101132113 in words is "one hundred one million, one hundred thirty-two thousand, one hundred thirteen".