The key take of Test case 3 is the order of checking if prime and palindrome.

prime?(x) && palindrome?(x) # does not pass
palindrome?(x) && prime?(x) # pass

It's quite clever to check the less complex operation first

**BY USING THIS CODE WE SOLVE THE ABOVE PROBLEM BUT THE THIRD CASE IS NOT COMPLETE ** def prime?(number) return ary = (2..number-1).collect{|i| i if number%i==0}.compact.length==0 end def palidrom?(num) i = 0 num = num.to_s j = num.length-1 while i<=j if(num[i]!=num[j]) return false end i+=1 j-=1 end return true end arr = [] num = gets.chomp.to_i i = 0 number = 2 while(i!=num)

if(palidrom?(number) && prime?(number))
i+=1
arr.push(number)
end
number+=1

end print arr

More elegant solution:

def prime?(x)
    return false if x < 2
    
    checker = Math.sqrt(x).to_i # 4
    2.upto(checker).each {|i| return false if x % i == 0 }
    
    true
end

palindrom_array = -> (array_size) do
    2.upto(Float::INFINITY).lazy.select do |x| 
        prime?(x) && x.digits.join.to_i == x
    end.first(array_size.to_i)
end

N = gets
print palindrom_array.(N)

Here is Lazy evaluation problem solution - https://www.gyangav.com/2022/10/hackerrank-lazy-evaluation-problem-solution.html