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by harsh.champ » Sun Feb 07, 2010 11:37 am
Let S be the set of integers 4, 12, 20, 28, ......., 516 and S' be the subset of S such that the sum of no two elements of S' is 520. The maximum possible number of elements in S' is

(A)30
(B)31
(C)32
(D)33
(E)35

The ans. is D.
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by shashank.ism » Sun Feb 07, 2010 11:44 am
harsh.champ wrote:Let S be the set of integers 4, 12, 20, 28, ......., 516 and S' be the subset of S such that the sum of no two elements of S' is 520. The maximum possible number of elements in S' is

(A)30
(B)31
(C)32
(D)33
(E)35

The ans. is D.
let n be the no. terms in set S.
The set is an A.P. with 1st term =4 and last term =516. and c.d. = 8
4+(n-1)8=516 --> n-1 = 512/8 =64 --> n=65.
so no. of terms in set S = 65

Now sum of two nos. which give 520 are(4+516, 12+508, 20+500, ......., 246+264) so 256 is left out.
Now if half of these nos. are there then no two terms will add to 520.
...
246= 4+(n-1)8 -->n-1 = 242/8 = 31 --> n=32
and 256 can also be added

so total element of subset S' = 32+1 =33
The ans. is D.[/quote]
Last edited by shashank.ism on Sun Feb 07, 2010 11:56 am, edited 1 time in total.
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by sars72 » Sun Feb 07, 2010 11:53 am
members of set s can be written in the form 8n-4, where n>=1

if no two numbers add upto 516, then there must be no number > 258

8*32 = 256
256-4 = 252

8*33 = 264
264-4 = 260

260+264 = 514

514< 516
--> 33 numbers

--> D
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