8 AS HIS OTHER PArt was on the table so only one possible to have 8 as 9 people and 9 possiblites was their from 0 to 8

he  put one A PILL AND PUT IN WATTER AND THEN DRINK HALF OF WATER

People will
view the biggest number they see, the person having 9 stares will realize he is biggest he will wear say black then after 10 seconds they will stare the second biggest they see.the person  having 8 stares will be second biggest  same continues ..and  yo yo who is  back .just Kidding .just a fighter mate.

using 11 division test we know abcdef if divisible by 11
a + c+e+(b+d+f)=2(sum of even or odd places) + some multiple of  11 having less digits then abcdef.now suppose statement in question works till n digits where n is any number then it will work for n+1 digits we just prove it works when n =2  so i did it

People will
view the biggest number they see, the person having 9 stares will realize he is biggest he will wear say black then after 10 seconds they will stare the second biggest they see.the person  having 8 stares will be second biggest  same continues ..and  yo yo who is  back .just Kidding .just a fighter mate

82709

7,8,9,10,11,12 on second dice.if 9 reverse is not considered as six

I will give 4 dollars and ask the waiter to keep the change

Make four groups of 3 piles and last 1 pile is kept away for while .now take 1 coin from every pile of 1 st group and 2 coins from every pile of 2 Nd group ……so on and put on the scale .

Expected weight is 30x divide the reading on the machine by 30 if the remainder is (25,30) or dividing by 30 remainder is (0,5) you get x in each case and now put back in original form then from different associated in 4th group take 1 from 1 pile and 3 from 1 and 4 from 1.

Similarly, from 3rd group take 1 coin from 1 pile and 2 coins from another and 4 from another .now from 1 St group 2 coin from 1 pile and 3 coins from another and 4 th from another from 2nd group take1 coin from 1 pile take 3 from another and 0 from 1 pile now put on the scale if no error then the pile 0 are chosen from have error rest is easy you just have to check the ratio of error to find out everything. now if remainder in first step is (20,25) or (5,10) for 30.you now x in each case,put back every thing in original form. Now from 1st group take 2 coin from 1 pile and 3 coin from 2nd pile and 4 coins from another pile.similarly from all other groups except in 4th one change is that the pile from which you were taking 2 coins.take. 4 coins instead and weigh rest is easy just observe the ratio of error and find everything you want.now if remainder is (10,15) it is clear that counterfeit coins are in the 3rd or 4th group .from 1st group take 2 coins from 1 pile and 3 from another and 4 from another.similarly from rest except 3rd group where you have to take 3 coins from pile where you have to take 4 if remainder remains same or decrease and is between (10,15) ten x is quotient if remainder increases then x is quotient +1 and rest is easy just use ratio of error to find your answer.if remainder is (15,20). Use the previous arrangement if remainder is same or increase you now x is quotient +1 else it is quotient and rest is easy use ratio of remainders .
Note I have not written too obvious things .i could not find close brackets so understand it yourself.if you could not understand once try again.
At last I deserve a giant present.finally relaxing time. Thanks.

He halved every pill and ate one half of every pill