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1104133502555 = 5376061798459
BaseRepresentation
bin10000000100010011011…
…111000011011001011011
310220112221221222110020222
4100010103133003121123
5121042231134040210
62203122032510255
7142525312630256
oct20042337033133
93815857873228
101104133502555
11396295639706
12159ba40ab38b
1380172270785
143b62445009d
151dac37e7655
hex101137c365b

1104133502555 has 16 divisors (see below), whose sum is σ = 1360806207840. Its totient is φ = 859410738432.

The previous prime is 1104133502509. The next prime is 1104133502611. The reversal of 1104133502555 is 5552053314011.

It is a cyclic number.

It is not a de Polignac number, because 1104133502555 - 214 = 1104133486171 is a prime.

It is a super-2 number, since 2×11041335025552 (a number of 25 digits) contains 22 as substring.

It is an unprimeable number.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 11164916 + ... + 11263374.

It is an arithmetic number, because the mean of its divisors is an integer number (85050387990).

Almost surely, 21104133502555 is an apocalyptic number.

1104133502555 is a deficient number, since it is larger than the sum of its proper divisors (256672705285).

1104133502555 is an equidigital number, since it uses as much as digits as its factorization.

1104133502555 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 159118.

The product of its (nonzero) digits is 45000, while the sum is 35.

Adding to 1104133502555 its reverse (5552053314011), we get a palindrome (6656186816566).

The spelling of 1104133502555 in words is "one trillion, one hundred four billion, one hundred thirty-three million, five hundred two thousand, five hundred fifty-five".

Divisors: 1 5 37 185 60617 98459 303085 492295 2242829 3642983 11214145 18214915 5968289203 29841446015 220826700511 1104133502555