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#CAT 17 QA Q. If you form a subset of integers chosen from between 1 to 3000, such that no two integers add up to a multiple of nine, what can be the maximum number of elements in the subset.

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numbers can be split into 1mod9,2mod9.. 0mod9(remainder when divided by 9). (1,8) (2,7) (3,6)(4,5) (0,0) make pairs which add upto 9. so only one of these pairs has to be chosen. Hence the answer is (333*4+3+1)= 1336
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Any number can be expressed of the form: 9K+i; where k= {0,1,2,....} and i={0,1,...,8} Now, we have to select numbers from {1,2,...,3000} in a manner so that final set does not contain any two numbers whose sum is divisible by 9. We have to identify the maximum size of such selection. Now, within 1 to... read more
Any number can be expressed of the form: 9K+i; where k= {0,1,2,....} and i={0,1,...,8} Now, we have to select numbers from {1,2,...,3000} in a manner so that final set does not contain any two numbers whose sum is divisible by 9. We have to identify the maximum size of such selection. Now, within 1 to 3000, #elements of the form 9k+1 = 334. #elements of the form 9k+2 = 334. #elements of the form 9k+3 = 334. #elements of the form 9k+4 = 333. #elements of the form 9k+5 = 333. #elements of the form 9k+6 = 333. #elements of the form 9k+7 = 333. #elements of the form 9k+8 = 333. #elements of the form 9k = 333. Now, as per condition, if you select any number from the group 9k+i then you can not select number from the group 9k+j where i+j=9. So, in order to maximize the size of selection, you will go for group: 9K+1, 9k+2,9K+3 and 9k+4. So, total (334+334+334+333=1335). Now, you can select exactly one number from the group 9K. (More than one will result their sum divisible by 9). So, total becomes 1336. Now, you have to exclude 1 (from group 9k+1) and 3000 (from group 9k+3). So finally you are with 1334 numbers! read less
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The question was meant for aspirants- I had written this in the post/ Admins do not chop off portions from the post when that is relevant/important.hope you understand.
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