Question 32297: What is the largest number of distinct postive intgers that can be found such that no four of them cab be selectedd with a sum divisable by 4?
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! WHEN A NUMBER IS DIVIDE BY 4 , WE CAN GET A REMAINDER OF 0..OR..1...OR...2..OR...3...
WE HAVE TO NAME INTGERS NO 4 OF WHICH WILL ADD UP TO A MULTIPLE OF 4.
SO LET US TAKE THE NUMBERS STARTING WITH THOSE GIVING A REMAINDER OF 0.
0...REMAINDER...WE CAN TAKE 3 SUCH NUMBERS..
1....REMAINDER....WE CAN TAKE THREE WITH THE ABOVE..AS ANY 4 TOGETHER WILL GIVE REMAINDER OF 3.OR..1..OR..2.
2....REMAINDER.......WE CAN NOT HAVE EVEN ONE WITH THE ABOVE...AS A SUM OF 0 REMAINDER CAN BE GOT FROM 4 OUT OF 3*0'S,3*1'S,1*2...BY TAKING...1*2+2*1+1*0
=4-4=0
3....REMAINDER .....WE CAN NOT HAVE EVEN ONE WITH THE ABOVE...AS A SUM OF 0 REMAINDER CAN BE GOT FROM 4 OUT OF 1*3,3*1,3*0...BY TAKING...1*3+1*1+2*0
=4-4=0
HENCE WE CAN HAVE A MAXIMUM OF 6 INTEGERS SAY
3 WITH REMAINDER 0 .....0,4,8,...
3 WITH REMAINDER OF 1....1,5,9.....
WE CAN CHECK THAT NO 4 OF THEM ADD UP TO A MULTIPLE OF 4...EX......
0+1+4+5=10
0+1+4+8=13
0+1+4+9=14
0+1+5+8=14
0+1+5+9=15
1+4+5+8=17
ETC...
...IF WE TRY TO ADD ANY NUMBER THEN WE CANT FULFILL THE REQUIREMENT ASKED.
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