Base | Representation |
---|---|
bin | 11110011000101110010100… |
… | …101110011001001111111011 |
3 | 122112011220020102202121111211 |
4 | 132120232110232121033323 |
5 | 120004031021031330201 |
6 | 1152121315311503551 |
7 | 40102125521246206 |
oct | 3630562456311773 |
9 | 575156212677454 |
10 | 133640402605051 |
11 | 396446382aa691 |
12 | 12ba44a4527bb7 |
13 | 597530b339084 |
14 | 250033118893d |
15 | 106b460b6c651 |
hex | 798b94b993fb |
133640402605051 has 2 divisors, whose sum is σ = 133640402605052. Its totient is φ = 133640402605050.
The previous prime is 133640402604977. The next prime is 133640402605057. The reversal of 133640402605051 is 150506204046331.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 133640402605051 - 29 = 133640402604539 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 133640402604995 and 133640402605013.
It is not a weakly prime, because it can be changed into another prime (133640402605057) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 66820201302525 + 66820201302526.
It is an arithmetic number, because the mean of its divisors is an integer number (66820201302526).
Almost surely, 2133640402605051 is an apocalyptic number.
133640402605051 is a deficient number, since it is larger than the sum of its proper divisors (1).
133640402605051 is an equidigital number, since it uses as much as digits as its factorization.
133640402605051 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 259200, while the sum is 40.
The spelling of 133640402605051 in words is "one hundred thirty-three trillion, six hundred forty billion, four hundred two million, six hundred five thousand, fifty-one".
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