Base | Representation |
---|---|
bin | 11111110111010101011010… |
… | …111011110100010001110010 |
3 | 200101012110000212202101221212 |
4 | 133313111122323310101302 |
5 | 121332041320431000000 |
6 | 1214020204025520122 |
7 | 41342632511264633 |
oct | 3767253273642162 |
9 | 611173025671855 |
10 | 140142013531250 |
11 | 407209921641a5 |
12 | 13874564943042 |
13 | 60274484a69b3 |
14 | 2686ca40d788a |
15 | 1130636d94c35 |
hex | 7f755aef4472 |
140142013531250 has 224 divisors, whose sum is σ = 278701232578560. Its totient is φ = 52786780200000.
The previous prime is 140142013531217. The next prime is 140142013531327. The reversal of 140142013531250 is 52135310241041.
140142013531250 is digitally balanced in base 3, because in such base it contains all the possibile digits an equal number of times.
It is an unprimeable number.
It is a pernicious number, because its binary representation contains a prime number (29) of ones.
It is a polite number, since it can be written in 111 ways as a sum of consecutive naturals, for example, 4231452191 + ... + 4231485309.
Almost surely, 2140142013531250 is an apocalyptic number.
140142013531250 is a gapful number since it is divisible by the number (10) formed by its first and last digit.
140142013531250 is a deficient number, since it is larger than the sum of its proper divisors (138559219047310).
140142013531250 is an frugal number, since it uses more digits than its factorization.
140142013531250 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 33308 (or 33283 counting only the distinct ones).
The product of its (nonzero) digits is 14400, while the sum is 32.
Adding to 140142013531250 its reverse (52135310241041), we get a palindrome (192277323772291).
The spelling of 140142013531250 in words is "one hundred forty trillion, one hundred forty-two billion, thirteen million, five hundred thirty-one thousand, two hundred fifty".
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