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11121021312000 = 21035328960993
BaseRepresentation
bin1010000111010101000010…
…0001100101110000000000
31110101011022002121022111020
42201311100201211300000
52424201322423441000
635352532054145440
72225316163311615
oct241652041456000
943334262538436
1011121021312000
1135a8442673936
1212b73b3652280
1362892809ac42
142a63903b650c
1514443b150ba0
hexa1d50865c00

11121021312000 has 176 divisors, whose sum is σ = 36992688544032. Its totient is φ = 2965605580800.

The previous prime is 11121021311999. The next prime is 11121021312013. The reversal of 11121021312000 is 21312012111.

It is a super-4 number, since 4×111210213120004 (a number of 53 digits) contains 4444 as substring.

It is a Harshad number since it is a multiple of its sum of digits (15).

It is a congruent number.

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, 14096497 + ... + 14864496.

Almost surely, 211121021312000 is an apocalyptic number.

11121021312000 is a gapful number since it is divisible by the number (10) formed by its first and last digit.

It is an amenable number.

11121021312000 is an abundant number, since it is smaller than the sum of its proper divisors (25871667232032).

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

11121021312000 is an equidigital number, since it uses as much as digits as its factorization.

11121021312000 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 28961031 (or 28961003 counting only the distinct ones).

The product of its (nonzero) digits is 24, while the sum is 15.

Adding to 11121021312000 its reverse (21312012111), we get a palindrome (11142333324111).

The spelling of 11121021312000 in words is "eleven trillion, one hundred twenty-one billion, twenty-one million, three hundred twelve thousand".