Blance NH3+Na--->NaNH2+H2

2NH3 + 2Na —> 2NaNH2 + H2

Explanation:

NH3 + Na —> NaNH2 + H2

The above can be balance by doing the following:

There are a total of 4 hydrogen atoms on the right and 3 atoms on the left. This can be balance by putting 2 in front of NH3 and 2 in front of NaNH2 as shown below:

2NH3 + Na —> 2NaNH2 + H2

There are 2 atoms of Na on the right and 1 atom on the left. It can be balance by putting 2 in front of Na as shown below

2NH3 + 2Na —> 2NaNH2 + H2

Now we can see that the equation is balanced as we have equal numbers of atoms of the different elements on both sides of the equation.

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