# How to create a list from 100 to 0 decrementing?

Posted 1 month ago
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 Hi there !! I should feel embarrassed to ask such a simple question but I can not find any answers on the web....maybe because it is too simple... I use Nestlist[ floor[]] as in the lines below, but if I increase the number it won't give an answer, is there a simpler way to list from 100 to 0 {100,99,98,97,96...,3,2,1,0}? or from n->0? n=1223 a=NestList[Floor[n--]&,n,n] 
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Posted 1 month ago
 Yes, there is a simple way. Range followed by Reverse Range[0, 100] // Reverse n = 1223 Range[n] // Reverse 
Posted 1 month ago
 Thank you so much Hans
Posted 1 month ago
 Or Table with a negative increment Table[x, {x, 100, 0, -1}] 
Posted 1 month ago
 Or  Range[100, 0, -1] 
Posted 1 month ago
 Thank you all but they all have a system memory failure issue, I don't know if divisors in reverse will reverse the calculus of the division or if it will only reverse the output...it is what I want to make the mod[a,n] n being the numbers down from n to 0... but for huge numbers until I find 0 for the result, a perfect divisor so to exclude the number from being a prime.
 Well if n is a huge integer then generating a list of {n, n - 1, n - 2, ... 0} is of course going to exhaust memory. You could try something like this p1000000 = Prime[1000000] (* Can optimize to only check from Sqrt[n] *) (n = p1000000; While[Mod[p1000000, n - 2] != 0, n = n - 2]; n - 2) // AbsoluteTiming (* {6.72053, 1} *) (n = p1000000 - 2; While[Mod[p1000000 - 2, n - 2] != 0, n = n - 2]; n - 2) // AbsoluteTiming (* {6.75608, 910933} *) Mod[p1000000 - 2, 910933] (* 0 *) If n is large, you are going to have to wait a long time. You could split the range and parallelize across available kernels, but still. You can probably estimate roughly how long it is going to take by computing the time to divide n by 1000 candidate divisors and extrapolating.