Base | Representation |
---|---|
bin | 100000000101101001000101… |
… | …011011111001100001001010 |
3 | 200111200102210212012220021110 |
4 | 200011221011123321201022 |
5 | 121444143402242301404 |
6 | 1220052004431351150 |
7 | 41503650011343420 |
oct | 4005510533714112 |
9 | 614612725186243 |
10 | 141125200353354 |
11 | 40a6a952522748 |
12 | 139b30140824b6 |
13 | 60990843a1259 |
14 | 26bc6d35b3110 |
15 | 114aecc538b89 |
hex | 805a456f984a |
141125200353354 has 32 divisors (see below), whose sum is σ = 322572448275840. Its totient is φ = 40321415596032.
The previous prime is 141125200353329. The next prime is 141125200353391. The reversal of 141125200353354 is 453353002521141.
141125200353354 is digitally balanced in base 3, because in such base it contains all the possibile digits an equal number of times.
It is a Curzon number.
It is an unprimeable number.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 24504082 + ... + 29710274.
It is an arithmetic number, because the mean of its divisors is an integer number (10080389008620).
Almost surely, 2141125200353354 is an apocalyptic number.
141125200353354 is a gapful number since it is divisible by the number (14) formed by its first and last digit.
141125200353354 is an abundant number, since it is smaller than the sum of its proper divisors (181447247922486).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
141125200353354 is a wasteful number, since it uses less digits than its factorization.
141125200353354 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 5851614.
The product of its (nonzero) digits is 216000, while the sum is 39.
Adding to 141125200353354 its reverse (453353002521141), we get a palindrome (594478202874495).
The spelling of 141125200353354 in words is "one hundred forty-one trillion, one hundred twenty-five billion, two hundred million, three hundred fifty-three thousand, three hundred fifty-four".
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