Answer - b)One - 211
Approach -
Step 1 - The list 200-220 has 11 even numbers and 10 odd numbers.
Step 2 - We are down to 10 possible-prime-numbers as the rest 11 are divisible by 2, hence non prime.
Step 3 - Out of the 10 possible-prime-numbers, 4 are divisible by 3 (201,207,213,219), hence non prime. so we are down to 6 possible-prime-numbers.
Step 4 - Out of the remaining 6 possible-prime-numbers, 2 are multiples of 5, hence non prime. Now, we are down to 4 possible-prime-numbers.
Step 5 - Now, the numbers 203, 209, 211, 217 are leftover. As the upper limit (200-220) is 220 and the nearest square is 225(15^2), check if the remaining 4 possible-prime-numbers are divisible by at least one among the prime numbers below 15(2,3,5,7,11,13,17,19).
203 and 217 are divisible by 7
209 is divisible by 11
So 211 is the only one remaining !
Looks and feels lengthy, but is really easy.(a 1 minute problem)
