Suppose we have 14 red balls and 14 green balls as in the previous exercise. Show that at least two pairs, consisting of one red and one green ball, have the same value. What about 13 red balls and 14 green balls

since each ball has a different number and if no two pairs have the same value there is going to be 14∗14 different sums. Looking at the numbers 1 through 100 the highest sum is 199 and lowest is 3, giving 197 possible sums

For the 14 case, we show that there exist at least one number from set {3,4,5,...,17} is not obtainable and at least one number from set {199,198,...,185} is not obtainable.

So we are left with 197 - 195 options

14 x 14 = 196

196 > 195

so there are two pairs consisting of one red and one green ball that have the same value

As to the comment, I constructed a counter-example list for the 13 case as follows. The idea of constructing this list is similar to the proof for the 14 case.

Red: (1,9,16,23,30,37,44,51,58,65,72,79,86)

Green: (2,3,4,5,6,7,8;94,95,96,97,98,99,100)

Note that 86+8=94 and 1+94=95 so there are no duplicated sum

Step-by-step explanation:

For the 14 case, we show that there exist at least one number from set {3,4,5,...,17} is not obtainable and at least one number from set {199,198,...,185} is not obtainable.

First consider the set {3,4,5,...,17}.

Suppose all numbers in this set are obtainable.

Then since 3 is obtainable, 1 and 2 are of different color. Then since 4 is obtainable, 1 and 3 are of different color. Now suppose 1 is of one color and 2,3,...,n−1 where n−1<17 are of the same color that is different from 1's color, then if n<17 in order for n+1 to be obtainable n and 1 must be of different color so 2,3,...,n are of same color. Hence by induction for all n<17, 2,3,...,n must be of same color. However this means there are 16−2+1=15 balls of the color contradiction.

Hence there exist at least one number in the set not obtainable.

We can use a similar argument to show if all elements in {199,198,...,185} are obtainable then 99,98,...,85 must all be of the same color which means there are 15 balls of the color contradiction so there are at least one number not obtainable as well.

Now we have only 195 choices left and 196>195 so identical sum must appear

A similar argument can be held for the case of 13 red balls and 14 green balls

mental cunning

explanation:

he is often referred to as a man of twists and turns

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.