The numbers can be grouped by pairs:
999,999,999 and 0;
999,999,998 and 1'
999,999,997 and 2;
and so on....
There are half a billion pairs, and the sum of the digits in each pair is 81. The digits in the unpaired number, 1,000,000,000, add to 1. Then:
(500,000,000 X 81) + 1= 40,500,000,001.
Fill the 5 oz. and 11 oz. Containers from the 24 oz. container. This leaves 8 oz. in the 24 oz. bottle. Next empty the 11 oz. bottle by pouring the contents into the 13 oz. bottle. Fill the 13 oz. bottle from the 5 oz. container (with 2 oz.) and put the remaining 3 oz. in the 11 oz. bottle. This leaves the 5 oz. container empty. Now pour 5 oz. from the 13 oz. bottle into the 5 oz. bottle leaving 8 oz. in the 13 oz. bottle. Finally pour the 5 oz. bottle contents into the 11 oz. bottle giving 8 oz. in this container.
Most people seem to think that the thing to do is weight six coins against six coins, but if you think about it, this would yield you no information concerning the whereabouts of the only different coin. As we already know that one side will be heavier than the other. So that the following plan can be followed, let us number the coins from 1 to 12. For the first weighing let us put on the left pan coins 1,2,3,4 and on the right pan coins 5,6,7,8. There are two possibilities. Either they balance, or they don't. If they balance, then the different coin is in the group 9,10,11,12. So for our second weighing we would put 1,2 in the left pan and 9,10 on the right. If these balance then the different coin is either 11 or 12. Weigh coin 1 against 11. If they balance, the different coin is number 12. If they do not balance, then 11 is the different coin. If 1,2 vs 9,10 do not balance, then the different coin is either 9 or 10. Again, weigh 1 against 9. If they balance, the different coin is number 10, otherwise it is number 9. That was the easy part. What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these coins could be the different coin. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings. Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the different coin is either 3 or 4. Weigh 4 against 9, a known good coin. If they balance then the different coin is 3, otherwise it is 4. Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a different, heavy coin, or 1 is a different, light coin. For the third weighing, weigh 7 against 8. Whichever side is heavy is the different coin. If they balance, then 1 is the different coin. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a different heavy coin or 2 is a light different coin. Weigh 5 against 6. The heavier one is the different coin. If they balance, then 2 is a different light coin.