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101113442432417 is a prime number
BaseRepresentation
bin10110111111011001001110…
…110100010111010110100001
3111021000100110122200121122202
4112333121032310113112201
5101223120312230314132
6555014512454142545
730204130235514662
oct2677311664272641
9437010418617582
10101113442432417
112a243a7425a648
12b410567351a55
134455c58b7022c
141ad7cb96d3369
15ba52d5468762
hex5bf64ed175a1

101113442432417 has 2 divisors, whose sum is σ = 101113442432418. Its totient is φ = 101113442432416.

The previous prime is 101113442432411. The next prime is 101113442432491. The reversal of 101113442432417 is 714234244311101.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 92210714174881 + 8902728257536 = 9602641^2 + 2983744^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-101113442432417 is a prime.

It is a super-2 number, since 2×1011134424324172 (a number of 29 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (101113442432411) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 50556721216208 + 50556721216209.

It is an arithmetic number, because the mean of its divisors is an integer number (50556721216209).

Almost surely, 2101113442432417 is an apocalyptic number.

It is an amenable number.

101113442432417 is a deficient number, since it is larger than the sum of its proper divisors (1).

101113442432417 is an equidigital number, since it uses as much as digits as its factorization.

101113442432417 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 64512, while the sum is 38.

Adding to 101113442432417 its reverse (714234244311101), we get a palindrome (815347686743518).

The spelling of 101113442432417 in words is "one hundred one trillion, one hundred thirteen billion, four hundred forty-two million, four hundred thirty-two thousand, four hundred seventeen".