#csirnet #mathematics FEB 2022 | Number Theory | number of elements in S^c

Hello friends, this is Mahesh Yuvalikar and you are watching NextGenMaths. If you have not subscribed my channel, then please subscribe the channel and press the bell icon. In this video, we are going to discuss the problem of CSI NET 5th 2022 Number Theory. Free Free Free Free Crash Course of CSI NET Mathematical Science is absolutely free on IFAS. Offer ends on 10th July 12 AM. So my friend, be fast. Register on IFAS app and off free access to an exclusive crash course made by India's finest. For more details, call on 9172266888 or click on the link which is available in the description. The problem is let S equal to N such that 1 less than equal to N less than equal to 999, 3 divides N or 37 divides N. How many integers are there in the set S complement equal to N such that 1 less than equal to N less than equal to 999 and does not belongs to S. We have gained 4 options. 1st option 639, 2nd option 648, 3rd option 668 and 4th option 990. Let's see the solution. Assume S equal to N such that 1 less than equal to N less than equal to 999, 3 divides N or 37 divides N. That means S contain numbers which are multiple of 3 or multiple of 37. Now we assume set A, N such that 1 less than equal to N less than equal to 999, 3 divides N. That means A contain the number which are multiple of 3. Assume set B, N such that 1 less than equal to N less than equal to 999, 37 divides N. That means B contain numbers which are multiple of 37. So S is a union of A and B. Number of element in A. A contain the numbers which are multiple of 3 and lies between 1 to 999. So we have 999 divide by 3. That means 333 numbers contain in A which are multiple of 3 between 1 and 999. Now number of element in B. B contain the number which are multiple of 37. So 999 divide by 37 equal to 27. That means 27 numbers contain in B. Now number of element in A intersection B. So A intersection B contain the numbers which are multiple of 3 as well as multiple of 37. Therefore A intersection B contain the numbers which are multiple of 3 and 37 both. So number of element in A intersection B is 999 divide by 37 into 3 which will give 9. That means 9 elements are in the set A intersection B. S equal to A union B. So number of element in S equal to number of element in A union B equal to set theoretical result number of A plus number of B minus number of A intersection B. So 333 plus 27 minus 9 it will give 351. So number of element in S is 351. Here we have to find the number of element in S complement which is n such that 1 less than equal to n less than equal to 999 and does not belongs to S. Therefore number of element in S complement equal to 999 minus number of element in S. So 999 minus 351 which will give 648 and which is a option number 2. Therefore option number 2 is the correct answer. Thanks for watching. For more such video like, share, subscribe, press the bell icon of next-gen maths.