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121300120113122 = 260650060056561
BaseRepresentation
bin11011100101001001100010…
…100100100000111111100010
3120220111011210221010212101202
4123211021202210200333202
5111344340121222104442
61105552301135001202
734356434024620556
oct3345114244407742
9526434727125352
10121300120113122
1135717101575145
1211730938969202
13528b72438a662
1421d4d58563866
15e05463da4c32
hex6e5262920fe2

121300120113122 has 4 divisors (see below), whose sum is σ = 181950180169686. Its totient is φ = 60650060056560.

The previous prime is 121300120113017. The next prime is 121300120113169. The reversal of 121300120113122 is 221311021003121.

It is a semiprime because it is the product of two primes.

It can be written as a sum of positive squares in only one way, i.e., 116426926997161 + 4873193115961 = 10790131^2 + 2207531^2 .

It is a junction number, because it is equal to n+sod(n) for n = 121300120113094 and 121300120113103.

It is an unprimeable number.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 30325030028279 + ... + 30325030028282.

Almost surely, 2121300120113122 is an apocalyptic number.

121300120113122 is a deficient number, since it is larger than the sum of its proper divisors (60650060056564).

121300120113122 is an equidigital number, since it uses as much as digits as its factorization.

121300120113122 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 60650060056563.

The product of its (nonzero) digits is 144, while the sum is 20.

Adding to 121300120113122 its reverse (221311021003121), we get a palindrome (342611141116243).

The spelling of 121300120113122 in words is "one hundred twenty-one trillion, three hundred billion, one hundred twenty million, one hundred thirteen thousand, one hundred twenty-two".

Divisors: 1 2 60650060056561 121300120113122