Step 1: we name all the bag of gold coins as #1, #2, #3......#8,
#9, and #10
Step 2: we put 1 coin from bag #1, 2 coins from bag #2,
3 coins from bag #3.........8 coins from bag #8, 9 coins
from bag #9, and 10 coins from bag 10 onto the scale.
Find out the total weight.
Step 3: the total weight should have been 10 grams X
(1+2+3+4+5+6+7+8+9+10=55) = 550 grams if all coins are the
same (10grams each).
Step 4: Subtract the total of step 2 from total of step
3.
Conclusion: If step 4 results 1 gram, then bag #1 is the
low quality coins, if step 4 results 2 grams, then bag
#2 is the one, if step 4 results 3 grams, then bag #3
is the one.......etc.
Thanks to Jason Vuong for this variation!
***Special Note: Another
visitor, Jay, offers this solution to this problem. Perhaps an easier way
to look at it?! Read on and see what you think:
1. I would put one coin on the scale from one of the bags.
2. If the change in weight on the scale is 10 grams, I 'd add a
coin from another bag.
3. I'll keep on adding a coin from each bag until the change in
the total weight after adding a coin is by 9 grams and the
bag that coin comes from is the low quality one.
I just made it look big and in 3 different steps to make it
sound easier but in practicality it's really simple and I
believe even a 3 year old kid can do it.
Thanks Jay!!
AND, ANOTHER VISITOR, Luis, POSES THIS:
I write you regarding weighty problem No.2 (gold coins)
I believe that Jay's solution to this problem does
not follow the restriction on the use of the scale.
The first coin you place on the scale will give a
reading, this will count
for one use of it, so you would not be able to
place another coin on it
because this will give you a second reading,
resulting in a second use of
the scale.
