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1553154000147 = 32237503159421
BaseRepresentation
bin10110100110011111001…
…100010100010100010011
312111110221222021010011200
4112212133030110110103
5200421324411001042
63145301544242243
7220132414114011
oct26463714242423
95443858233150
101553154000147
11549763a40432
12211018170383
13b360073a3b9
145525cc0d8b1
152a603aa2c4c
hex1699f314513

1553154000147 has 12 divisors (see below), whose sum is σ = 2340985739664. Its totient is φ = 990417043440.

The previous prime is 1553154000127. The next prime is 1553154000161. The reversal of 1553154000147 is 7410004513551.

It is not a de Polignac number, because 1553154000147 - 216 = 1553153934611 is a prime.

It is a super-3 number, since 3×15531540001473 (a number of 38 digits) contains 333 as substring.

It is not an unprimeable number, because it can be changed into a prime (1553154000127) by changing a digit.

It is a polite number, since it can be written in 11 ways as a sum of consecutive naturals, for example, 3751579504 + ... + 3751579917.

It is an arithmetic number, because the mean of its divisors is an integer number (195082144972).

Almost surely, 21553154000147 is an apocalyptic number.

1553154000147 is a deficient number, since it is larger than the sum of its proper divisors (787831739517).

1553154000147 is a wasteful number, since it uses less digits than its factorization.

1553154000147 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 7503159450 (or 7503159447 counting only the distinct ones).

The product of its (nonzero) digits is 42000, while the sum is 36.

Adding to 1553154000147 its reverse (7410004513551), we get a palindrome (8963158513698).

The spelling of 1553154000147 in words is "one trillion, five hundred fifty-three billion, one hundred fifty-four million, one hundred forty-seven", and thus it is an aban number.

Divisors: 1 3 9 23 69 207 7503159421 22509478263 67528434789 172572666683 517718000049 1553154000147