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121120000220112 = 24329241131558629
BaseRepresentation
bin11011100010100001110010…
…100101101001101111010000
3120212211221220001210020212020
4123202201302211221233100
5111333412230024020422
61105333432052331440
734340426342033265
oct3342416245515720
9525757801706766
10121120000220112
1135657781594633
1211701a4b013b80
13527774a958aca
1421ca34c87926c
15e00920db795c
hex6e2872969bd0

121120000220112 has 240 divisors, whose sum is σ = 334493217938880. Its totient is φ = 37740013875200.

The previous prime is 121120000220081. The next prime is 121120000220123. The reversal of 121120000220112 is 211022000021121.

It is a junction number, because it is equal to n+sod(n) for n = 121120000220091 and 121120000220100.

It is a congruent number.

It is an unprimeable number.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written in 47 ways as a sum of consecutive naturals, for example, 216537214 + ... + 217095842.

It is an arithmetic number, because the mean of its divisors is an integer number (1393721741412).

Almost surely, 2121120000220112 is an apocalyptic number.

121120000220112 is a gapful number since it is divisible by the number (12) 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 121120000220112, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (167246608969440).

121120000220112 is an abundant number, since it is smaller than the sum of its proper divisors (213373217718768).

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

121120000220112 is a wasteful number, since it uses less digits than its factorization.

121120000220112 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 558870 (or 558835 counting only the distinct ones).

The product of its (nonzero) digits is 32, while the sum is 15.

Adding to 121120000220112 its reverse (211022000021121), we get a palindrome (332142000241233).

The spelling of 121120000220112 in words is "one hundred twenty-one trillion, one hundred twenty billion, two hundred twenty thousand, one hundred twelve".