# Math Help - Arbitary Arrangement

1. ## Arbitary Arrangement

I need help again

Question is as follows:

Twelve basketball players whose uniforms are numbered 1 through 12 stand in a line in an arbitrary arrangement. prove that there must be three consecutive players in the line such that the sum of the numbers of their uniforms is at least 20.

2. Originally Posted by faisalnet5
I need help again

Question is as follows:

Twelve basketball players whose uniforms are numbered 1 through 12 stand in a line in an arbitrary arrangement. prove that there must be three consecutive players in the line such that the sum of the numbers of their uniforms is at least 20.

$1+2+..+12=78$
considering the (1,2,3) (4,5,6)(7,8,9)(10,11,12),four 3-tuples.((1,2,3)means the consecutive series formed by the first,second,third people)
since $78\div 4=19.5$ and the sum of the numbers in (1,2,3),(4,5,6),(7,8,9),(10,11,12) is 78, so one one them must be greater than 19.5, so greater than 20.

3. But Yuj....how did u get 156\div 4=19.5?

4. I think i still couldn't get the question....can any one please help me to solve?

Regards

5. Originally Posted by faisalnet5
I think i still couldn't get the question....can any one please help me to solve?

Regards
there is a mistake.... $1+2+..+12=78$...sorry

6. Hello faisalnet5
Originally Posted by faisalnet5
I think i still couldn't get the question....can any one please help me to solve?

Regards
Imagine each number from 1 to 12 being represented by 1 bead, 2 beads, ... 12 beads. The total number of beads is therefore 1 + 2 + ... + 12 = 78.

Now imagine the players standing in a line, and you ask them to get into groups of three, starting at one end of the line and ending at the other. Now imagine each group of three to be represented by a box. In each box, then, we imagine a number of beads - the total number represented by the numbers on the players' backs.

We have to show that at least one box contains 20 or more beads.

So, we have 78 beads to put into four boxes. If we put 19 into each box - that's 19 x 4 = 76 beads. There are still 2 beads left over. So?