Base | Representation |
---|---|
bin | 11110001111100100010101… |
… | …010110100111111010100100 |
3 | 122102221202011020212222012212 |
4 | 132033210111112213322210 |
5 | 114413223414303121040 |
6 | 1150520300225502552 |
7 | 40005512354146505 |
oct | 3617442526477244 |
9 | 572852136788185 |
10 | 133011200442020 |
11 | 3942180806685a |
12 | 12b02565aa1a58 |
13 | 592ab966bcc74 |
14 | 24bbac2820bac |
15 | 1059dd7271d65 |
hex | 78f9155a7ea4 |
133011200442020 has 12 divisors (see below), whose sum is σ = 279323520928284. Its totient is φ = 53204480176800.
The previous prime is 133011200442019. The next prime is 133011200442023. The reversal of 133011200442020 is 20244002110331.
It can be written as a sum of positive squares in 2 ways, for example, as 14129118712384 + 118882081729636 = 3758872^2 + 10903306^2 .
It is a super-3 number, since 3×1330112004420203 (a number of 43 digits) contains 333 as substring.
It is not an unprimeable number, because it can be changed into a prime (133011200442023) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 3325280011031 + ... + 3325280011070.
It is an arithmetic number, because the mean of its divisors is an integer number (23276960077357).
Almost surely, 2133011200442020 is an apocalyptic number.
133011200442020 is a gapful number since it is divisible by the number (10) formed by its first and last digit.
It is an amenable number.
133011200442020 is an abundant number, since it is smaller than the sum of its proper divisors (146312320486264).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
133011200442020 is a wasteful number, since it uses less digits than its factorization.
133011200442020 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 6650560022110 (or 6650560022108 counting only the distinct ones).
The product of its (nonzero) digits is 1152, while the sum is 23.
Adding to 133011200442020 its reverse (20244002110331), we get a palindrome (153255202552351).
The spelling of 133011200442020 in words is "one hundred thirty-three trillion, eleven billion, two hundred million, four hundred forty-two thousand, twenty".
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