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2000131554451 = 49311541326317
BaseRepresentation
bin11101000110110001001…
…000010111110010010011
321002012200121101111202011
4131012301020113302103
5230232232134220301
64130503040403351
7264335060461522
oct35066110276223
97065617344664
102000131554451
11701284134678
12283780016557
131167c48a5544
146cb421364b9
1537064816a51
hex1d1b1217c93

2000131554451 has 8 divisors (see below), whose sum is σ = 2000742995664. Its totient is φ = 1999520206560.

The previous prime is 2000131554433. The next prime is 2000131554461. The reversal of 2000131554451 is 1544551310002.

It is a sphenic number, since it is the product of 3 distinct primes.

It is a cyclic number.

It is not a de Polignac number, because 2000131554451 - 25 = 2000131554419 is a prime.

It is a super-2 number, since 2×20001315544512 (a number of 25 digits) contains 22 as substring.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (2000131554401) by changing a digit.

It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 75988345 + ... + 76014661.

It is an arithmetic number, because the mean of its divisors is an integer number (250092874458).

Almost surely, 22000131554451 is an apocalyptic number.

2000131554451 is a deficient number, since it is larger than the sum of its proper divisors (611441213).

2000131554451 is a wasteful number, since it uses less digits than its factorization.

2000131554451 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 46661.

The product of its (nonzero) digits is 12000, while the sum is 31.

Adding to 2000131554451 its reverse (1544551310002), we get a palindrome (3544682864453).

The spelling of 2000131554451 in words is "two trillion, one hundred thirty-one million, five hundred fifty-four thousand, four hundred fifty-one".

Divisors: 1 4931 15413 26317 76001503 129769127 405623921 2000131554451