Base | Representation |
---|---|
bin | 11001011111100100110101… |
… | …000101111010000100110001 |
3 | 112200222122221002211022212200 |
4 | 121133210311011322010301 |
5 | 104143442313033044441 |
6 | 1034243403140013413 |
7 | 32421322125212502 |
oct | 3137446505720461 |
9 | 480878832738780 |
10 | 112121012003121 |
11 | 327a8288318010 |
12 | 106a9978091269 |
13 | 4a73c735a05b4 |
14 | 1d989862caba9 |
15 | ce67d14487b6 |
hex | 65f93517a131 |
112121012003121 has 48 divisors (see below), whose sum is σ = 176902117337088. Its totient is φ = 67865031554400.
The previous prime is 112121012003111. The next prime is 112121012003129. The reversal of 112121012003121 is 121300210121211.
It is a happy number.
112121012003121 is a `hidden beast` number, since 1 + 1 + 21 + 210 + 120 + 0 + 312 + 1 = 666.
It is not a de Polignac number, because 112121012003121 - 29 = 112121012002609 is a prime.
It is a super-2 number, since 2×1121210120031212 (a number of 29 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 112121012003094 and 112121012003103.
It is not an unprimeable number, because it can be changed into a prime (112121012003129) by changing a digit.
It is a polite number, since it can be written in 47 ways as a sum of consecutive naturals, for example, 268543636 + ... + 268960826.
It is an arithmetic number, because the mean of its divisors is an integer number (3685460777856).
Almost surely, 2112121012003121 is an apocalyptic number.
112121012003121 is a gapful number since it is divisible by the number (11) formed by its first and last digit.
It is an amenable number.
112121012003121 is a deficient number, since it is larger than the sum of its proper divisors (64781105333967).
112121012003121 is a wasteful number, since it uses less digits than its factorization.
112121012003121 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 420682 (or 420679 counting only the distinct ones).
The product of its (nonzero) digits is 48, while the sum is 18.
Adding to 112121012003121 its reverse (121300210121211), we get a palindrome (233421222124332).
The spelling of 112121012003121 in words is "one hundred twelve trillion, one hundred twenty-one billion, twelve million, three thousand, one hundred twenty-one".
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