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1355054404115 = 51729549717811
BaseRepresentation
bin10011101101111111100…
…010010101001000010011
311210112122001220221211102
4103231333202111020103
5134200122411412430
62514300422335015
7166620340251236
oct23557742251023
94715561827742
101355054404115
11482747980925
1219a750b8a46b
139aa20063919
144982929d41d
15253ac3bb345
hex13b7f895213

1355054404115 has 16 divisors (see below), whose sum is σ = 1781085710880. Its totient is φ = 985094315520.

The previous prime is 1355054404073. The next prime is 1355054404157. The reversal of 1355054404115 is 5114044505531.

It is an interprime number because it is at equal distance from previous prime (1355054404073) and next prime (1355054404157).

It is a de Polignac number, because none of the positive numbers 2k-1355054404115 is a prime.

It is a super-2 number, since 2×13550544041152 (a number of 25 digits) contains 22 as substring.

It is an unprimeable number.

It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 274856441 + ... + 274861370.

It is an arithmetic number, because the mean of its divisors is an integer number (111317856930).

Almost surely, 21355054404115 is an apocalyptic number.

1355054404115 is a deficient number, since it is larger than the sum of its proper divisors (426031306765).

1355054404115 is a wasteful number, since it uses less digits than its factorization.

1355054404115 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 549717862.

The product of its (nonzero) digits is 120000, while the sum is 38.

Adding to 1355054404115 its reverse (5114044505531), we get a palindrome (6469098909646).

The spelling of 1355054404115 in words is "one trillion, three hundred fifty-five billion, fifty-four million, four hundred four thousand, one hundred fifteen".

Divisors: 1 5 17 29 85 145 493 2465 549717811 2748589055 9345202787 15941816519 46726013935 79709082595 271010880823 1355054404115