CSIR NET MATHEMATICS| Number Theory | Without Theory how to solve the problem of Number Theory?

In this video, we are going to discuss the problem of number theory from CSIR NET December 2018. Example 90, booklet C, which is of mark 4.75. Let a1 less than a2 less than a51 be given distinct natural numbers such that 1 less than equal to ai less than equal to 100 for all i from 1 to 51. Then, which of the following are correct? First option, there exist i and j with 1 less than equal to i less than equal to j less than equal to 51. Satisfying, a i divides a j. There exist i with 1 less than equal to i less than equal to 51 such that ai is an odd number. That means in the sequence, we can find at least one odd number. The third option, there exist... j with 1 less than equal to j less than equal to 51 such that aj is an even number that means sequence contains at least one even number there exists ai less than aj such that modulus ai minus aj is greater than 51 that we have given 51 distinct natural numbers between 1 to 100 with a1 less than a2 less than a51. What are the base possibility for the selection of this 51 number? So first possibility, selection of consecutive numbers. That means 1, 2, 3, 4, 51. You can take 2, 3, 4, 52, 3, 4, 54. Similarly, you can take 50, 51, 52, 53, 54, 100. So these are the possibility of selection of 51 numbers. The other possibility is random number selection that means 2, 4, 5, 10, 11. So we can select 51 numbers from 1 to 100 in any sequence but it should satisfy the condition a1 less than a2 less than 51. Now first option. What is the first option? There exists i and j with 1 less than equal to i less than equal to j less than equal to 51 satisfying ai divides aj. That means we are able to find at least one i and j such that a i divides a j. That means a j is a multiple of a i. So first we try to make a sequence of numbers such that it contains no multiple numbers and we satisfy the condition a1 less than a2 less than 51. So best possibility for that is 51 less than 52 less than 53 less than 54 less than 100. So this is the base possibility of selection of numbers where we are not getting any multiple numbers. These are only the 50 numbers and we want 51 numbers. So we must add the remaining number from 1 to 50 because selection of distinct 51 numbers. So the remaining one number from 1 to 50. It is a 50 number number 50 is selected as a 51 number then we have 100 which is a multiple of 50 numbers now if we had 49 then we have a 98 which is a multiple of 49 suppose we had 32 then 64 is there if it is 15 then 60 is there 7 then 63 if 1 then every number is a multiple of 1 so we observe We are able to find at least one number which is a multiple of the 51th number therefore the first option is correct option. Now for the second option, what is the second option? There exists i with 1 less than equal to i less than equal to 51 such that ai is an odd integer that means in the sequence there exists at least one odd number odd integer. Suppose we begin with all even numbers So 2, 4, 6, 8 and the 100 similarly this sequence of even numbers contain only 50 numbers and we want the 51 number so we should add at least one number and that number must be the odd number so we can add 1, we can add 3, we can add 37 but it must be the one odd number and therefore The second option is also correct. With the similar argument we say there exists at least one even number so we begin with the odd number sequence 1 2 99 these are only the 50 numbers so we have to add one number for 51 numbers and that number must be the even number it may be 2 it may be 8 it may be the 46 therefore The third option is also correct option. We are going to discuss the fourth option. Now fourth option is there exists a less than j such that modulus a i minus a j greater than 51. If we are able to find one sequence of 50 number which does not satisfy this condition then it is a wrong. Now if we begin with the simplest sequence 1 to 51 consecutive numbers then highest difference between a i and a j is A1 minus A51. So, 51 minus 1 which is 50 and which is not greater than 51. So, we cannot find any other pair because it is the highest difference. So, this sequence does not satisfy the fourth condition or we are not able to find the pair Ai, Aj such that modulus Ai minus Aj greater than 51. Therefore, fourth option is wrong. So, this Example has three correct option 1, 2, 3. Thanks for watching. For more videos, like, subscribe, press the bell icon of NextGenMS.