1Subho I thinku r wrong here :
the order is :
1,2,3 7,8,9,10 15,16,17 22,23,24 28,29,30,31,.....
which is much complicated.
I believe there's a simpler technique
Let X= {1,2,...,100} & Y be a subset of X such thatv the sum of no two elements in Y is
divisible by 7.If the max possible no of elements in Y is 40 + λ then λ is ---------??
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UP 0 DOWN 0 0 5
5 Answers
49thhe sum must not be 7,14,21,35,42,49,.....
if 1 is a member, 6,13,20,27,34,....rejected
if 2 is a member 5,12,19,26,33,..... "
" 3 " " '' 4,....... "
next member is 7,8,9,10
if we can see some order,
0,1,2,3 7,8,9,10 ... ... gives 14,15,16,17 21,22,23,24 so upto 91 plus 98,99,100
ans=4*14-1+3=40+18
please verify...
1
62we can chose only one number of the form 7k+1 or 7k+6
either of the form 7k+2 or 7k+5
either of the form 7k+3 or 7k+4
and 7k
what is interesting is that we can still chose one more number of the form 7k+1
so we will have
one nu;mber of the form 7k , 15 of 7k+1, 15 of 7k+2, 14 of 7k+3
hence the total numbers is 45