A set S consists of 43 natural numbers each of which is a perfect cube. The maximum no of elements of S that one can always find such that each of them leaves the same remainder when divided by 13 is? Ans-29.. approach please

1^3=1
2^3=8
3^3=1
4^3=12
5^3=8
6^3=8
7^3=5
8^3=5
9^3=1
10^3=12
11^3=5
12^3=12
13^3=0
again there is a repeat among remainders from 14^3 =1 same as 1^3 upto 26^3 =0 the same whole set
and again keep on eriting the same remainders from 27^3 to 39^3
then again do the same from 40^3 to 43^3 ul get remainders of all 43 first natural cubes .
Then count the repeated most repeated ones.
My apologies for any mistakes in the calculations and I assumed those are first 43 natural cubes .
I dint get 29 sry for that but i know how to do reminders using rem theory . Assuming im right .

but as it is not mentioned consecutive first 43 cubes v can use all such cubes which will give the same remainder when divided by 13
i.e. 1^3 , 3^3 , 9^3 , 14^3, 16^3, 22^3 ....( rem. is 1 in all the cases ) I think the ans is as simple as this because no exam will give such a lengthy qn
Ans - 43