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100613133553 = 211691475283
BaseRepresentation
bin101110110110100000…
…0101001010011110001
3100121200220100202100211
41131231000221103301
53122023430233203
6114115424344121
710161200344612
oct1355500512361
9317626322324
10100613133553
1139740544449
12175bb175041
13964585626a
144c26665c09
15293ce9dc6d
hex176d0294f1

100613133553 has 4 divisors (see below), whose sum is σ = 100613820528. Its totient is φ = 100612446580.

The previous prime is 100613133521. The next prime is 100613133563. The reversal of 100613133553 is 355331316001.

It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4, and also a brilliant number, because the two primes have the same length.

It is a cyclic number.

It is not a de Polignac number, because 100613133553 - 25 = 100613133521 is a prime.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (100613133563) by changing a digit.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 25951 + ... + 449332.

It is an arithmetic number, because the mean of its divisors is an integer number (25153455132).

Almost surely, 2100613133553 is an apocalyptic number.

It is an amenable number.

100613133553 is a deficient number, since it is larger than the sum of its proper divisors (686975).

100613133553 is an equidigital number, since it uses as much as digits as its factorization.

100613133553 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 686974.

The product of its (nonzero) digits is 12150, while the sum is 31.

Adding to 100613133553 its reverse (355331316001), we get a palindrome (455944449554).

The spelling of 100613133553 in words is "one hundred billion, six hundred thirteen million, one hundred thirty-three thousand, five hundred fifty-three".

Divisors: 1 211691 475283 100613133553