Question

I am trying to find some special numbers such that when it is divided by 2 gives 1, divided by 3 gives 2 and so on upto 6.

This perfectly works.

[ x | x <- [1..1000],x `mod` 2 == 1 , x `mod` 3 == 2 , x `mod` 4 == 3 , x `mod` 5 == 4 , x `mod` 6 == 5]

Ans:

[59,119,179,239,299,359,419,479,539,599,659,719,779,839,899,959]

I am trying to make it better so that it is not too verbose but the following doesn’t work.

[ x | x <- [1..1000], y <- [2..6], x `mod` y == (y-1) ]

It takes all x for which any of the y satisfies the condition but what I want is, I want x which satisfies the condition for all y.

Answer 1

You can write the condition as

[ x | x <- [1..1000], all (\y -> x `mod` y == y-1) [2 .. 6]]

But you can do better in this specific case,

let modulus = foldl1 lcm [2 .. 6]
[x | x <- [1 .. 1000], x `mod` modulus == modulus - 1]
[modulus - 1, 2*modulus - 1 .. 1000]  -- even better