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41201280000000 = 213325723311
BaseRepresentation
bin10010101111000111010111…
…01010001010000000000000
312101212210201001011212012100
421113203223222022000000
520400020210140000000
6223343335015430400
711451460061421141
oct1127435352120000
9171783631155170
1041201280000000
11121454014990aa
124755101385400
1319cb358107003
14a2621c9d9ac8
154b6b16e96a00
hex2578eba8a000

41201280000000 has 1344 divisors, whose sum is σ = 155740495053312. Its totient is φ = 10475520000000.

The previous prime is 41201279999989. The next prime is 41201280000073. The reversal of 41201280000000 is 8210214.

It is a Harshad number since it is a multiple of its sum of digits (18).

It is an unprimeable number.

It is a polite number, since it can be written in 95 ways as a sum of consecutive naturals, for example, 132479999845 + ... + 132480000155.

Almost surely, 241201280000000 is an apocalyptic number.

41201280000000 is a gapful number since it is divisible by the number (40) formed by its first and last digit.

It is an amenable number.

It is a practical number, because each smaller number is the sum of distinct divisors of 41201280000000, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (77870247526656).

41201280000000 is an abundant number, since it is smaller than the sum of its proper divisors (114539215053312).

It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.

41201280000000 is an frugal number, since it uses more digits than its factorization.

41201280000000 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 401 (or 344 counting only the distinct ones).

The product of its (nonzero) digits is 128, while the sum is 18.

Adding to 41201280000000 its reverse (8210214), we get a palindrome (41201288210214).

The spelling of 41201280000000 in words is "forty-one trillion, two hundred one billion, two hundred eighty million", and thus it is an aban number.