Let m be my age in years. If s is my son's age in years, then my son is 52s weeks old. If g is my grandson's age in years, then my grandson is 365g days old. Thus,

365g = 52s.

Since my grandson is 12g months old,

12g = m.

Since my grandson, my son and I together are 120 years,

g + s + m = 120.

The above system of 3 equations in 3 unknowns (g, s and m) can be solved as follows.

m / 12 + 365 m / (52 x 12) + m = 120 or

52 m + 365 m + 624 m = 624 x 120 or

m = 624 x 120 / 1041 = 72.

So, I am 72 years old.

An alternative solution is possible on realizing the significance of the word "about" in the Problem Statement.

The first equation (365g = 52s) can be approximated by

7g = s.

As before, the other two equations are

12g = m

g + s + m = 120.

The above system of 3 equations in 3 unknowns (g, s and m) can be simply solved as follows.

g + 7g + 12g = 120 or 20g = 120.

m = 12g = 12 x 120 / 20 = 72.

So, Grandpa is 72 years old.

A jail warden decides to throw a bit of a party for the staff and orders 1000 bottles of beer. As luck would have it, the warden is informed that one of the bottles is poisened and kills within 24 hours after only drinking the tiniest amount of the poisoned beer. No problem thinks the warden - I'll get some prisoners to do some quick tasting to find the poisoned bottle.

You pour a little of the first 500 bottles in beerpull #1 and a little of each in #2. One of them dies, other one survives. Then 2nd test 250 little parts of it in beerpull #3 and a little of those 250 in #4. you can use the surviving prisoner of the previous test, so only need 1 new.

500 each : 2 new prisoners
250 each : 1 new prisoner
125 idem
64 idem
32 idem
16 idem
8 idem
4 idem
2 idem

Makes 10 prisoners total.

NOTE: of course 125/ 2 is NOT 64, but then mix of the 62 or 63 left to be tested with a little of the earlier tested free-poison beer.

With the halving method, absurdly, you may have to taste 1001 beer samples! But you're guaranteed to taste at least 999, depending on which of the 62/63 and 31/32 batches proves poisoned. A waste of beer, if you ask me. You may have to take 10 sips of the same bottle. Also, anywhere between 0 and 10 prisoners may die.

Of course, if you have 999 inmates handy and the wardens are really thirsty, you can just go ahead and give every prisoner one bottle. The result will be known after one cycle, and you'll probably have 1 corpse on your hands. Quickest, simplest solution, pretty much the same as #2, only accelerated at the expense of wasting a lot of beer.

So, IMHO, there is no definitive answer to the problem. 1 prisoner is enough, but it'll be slow then. If you want an answer asap, then 999. If you want an answer reasonably fast, then anywhere from 1 to 10.

Give 20 prisoners 50 bottles each, and you'll be have 980 certified drinkable bottles within one cycle, and you only need those 50 guys, as opposed to 999, with the brute force method. You can use any number of methods to narrow that 20 down to 1.

Or you can go about thirding the bottles every cycle. Give one guy samples from 334, another one from 333. If neither dies, you go on dividing the remaining 333. Next cycle: 112, 111. Next: 38, 37 etc. This will be quicker than the halving method. My calculation suggests 7 cycles, maximum of 7 deaths. Maximum of 7 inmates necessary, 2 at a time. Around 2333 (!!!) samplings, and around 2328 even in the luckyest case.

Plus, of course, the halving method does NOT deliver an answer in one cycle at all. In each round, you have to wait to see which half to test in the next cycle. So it'll take 10 days.

<dakgootje> I misinterpreted your answer - but you still get credit for having the correct idea if not the excecution.

Line up the ten prisoners and label the bottles number 1,2,3....1000 etc. Now every number has a UNIQUE 10 digit binary repesentation. 1 is 0000000001 and 2 is 0000000010 and 3 is 0000000011 ... and 1000 is 1000011000. So we have each prisoner correspond to a binary digit. So with bottle one 0000000001 the last prisoner is the only to sample it. For bottle two 0000000010 the second last prisoner samples it. Continue this for all 1000 bottles. The combination of prisoners that eventually die is the binary expansion of the poisened bottle.

<You have a set of weighing scales and can pick four weights of whole number values. Which weights should you pick so that you can measure the largest consecutive unit weights starting from one unit? The problem is none too difficult, but I wonder if you can work out the largest number of consecutive unit weights that can be measured using n weights. I know a kind of solution to this extension, but have not studied maths at a high enough level (degree) to fully give an answer.>Posted by <kellmano>

<A king has twelve gold pawns-one pawn is lighter or heavier than the other eleven gold pawns.The king gives the wise man three weighings on a balance beam scale to tell him which pawn is lighter or heavier than the rest.1#The wise man can put any amount of pawns on each side of the scale and look if one side goes down then one side goes up(that is one weighing) (if the scale balances that's a weighing)The wise man says if i can make (THREE TOTAL WEIGHINGS) I will tell you which golden pawn is lighter or heavier than the other eleven golden pawns:The king offers his daughter and a room full of gold if the wise man can tell him.SCALE=

--------|-------->Posted by <capatal>

Both are solvable using base 3 maths.

The second one, the odd pawn, you dont know whether it is lighter of heavier, right?

The trick is to choose your three weighings such that every possible combination of results will specify a single unambiguos answer. Here is where the base 3 math comes. Every number between 1 and 12 can be expanded as a sum or difference of powers of 3. So 1 can be written as 3º, 2 can be written as -3º+3¹,...., and 12 can be written as 3¹+3². (note that since 13 can be expressed as powers of 3 not greater than 3² you do it for 13 if you had a good pawn to even out the weighings.).

So we have 1=3º,2=-3º+3¹,3=3¹,4=3º+3¹,5=-3º-3¹+3²,6=-3¹+3²,7=3º-3- ¹+3²,8=-3º+3²,9=3²,10=3º+3²,11=-3º+3¹+3²,12=3¹+3².

Why do we need these? The idea is that we assign one side of the scale to be negative and the other side to be positive. We use the sign of 3º in each numbers expansion to determine where to place it on the scale for the first weighing, the sign of 3¹ for the second weighing and finally the sign of 3² for the third weighing. The use of base 3 in this way gaurentees that such a method will generates weighings giving a solution. There is one small snag though - there are 8 +3²'s and no -3²'s. So we need to negate the numbers in such a way to leave equal numbers of positve and negative powers.

One such way is 1=3º,2=-3º+3¹,3=3¹,4=3º+3¹,5=-3º-3¹+3²,-6=3¹-3²,7=3º-3- ¹+3²,8=-3º+3²,9=3²,-10=-3º-3²,-11=3º-3¹-3²,-12=-3¹-3². This gives the weighings (stating the positive side of the scale first) as follows: First weighing->1,4,7,11 v 2,5,8,10. Second weighing->2,3,4,6 v 5,7,11,12. Third weighing->5,7,8,9 v 6,10,11,12.