Find the conterfeit coin?
If there was a conterfeit penny in a batch of ten pennies (the counterfeit is either lighter or heavier than the other coins) what is the least amount of weighings it would take to find the counterfeit?
Related posts:
- find the conterfeit coin? If there was a conterfeit penny in a batch of ten pennies (the counterfeit is either lighter or heavier than the other coins) what is the least amount of weighings...
- Of 80 coins of the same denomination, one is conterfeit and lighter than the others. ? Of 80 coins of the same denomination, one is counterfeit and lighter than the others. Identify the counterfeit coin by using a pan balance in no more than four weighings?...
- can you conterfeit coins?????????????????????????????????????????????????????????????????????????????????? People say you can’t but if you can counterfeit paper money why not coins????????????????????...
- Coins Coins and 12 Coins? Suppose you have 12 coins, one of which is counterfiet ( that is, it is either heavier or lighter). Using a balance, determine in three weighings, which coin is the...
- Hard MATH PROBLEM HELP PLEASSE!!!!!!!!!????? THere are twelve coins that are numbered 1 through 12. Eeleven weigh the same and one is either lighter or heavier than the others. Using just three weighings with a...
3, you need to weigh 2 of the pennies to find out how much the normal pennies weigh. If you find one that doesn’t match the other 2 in that set, then that one is the counterfeit.
Four weighings IS THE MAX necessary. — assuming you have a scale to put them on one side and the other side of the scale. And assuming you know the exact weight of one penny, so you can set a digital scale to what five pennies should weigh.
First weighing, you put on the scale 5 pennies. If the weight is either higher or lower than what it should be, then you know the "bad" penny is in that group.
f that weighing is equal to right amount, you know the "bad" penny is in the unweighed group of 5 pennies..
You divide the last five into two coins, and three coins.
Weigh the first two. If that weight is not true, to what two should weigh, you know one of those two is "bad".
If the weight is true, you know the bad penny is one of the other three.
In the third weighing, you then select two out of the other three, and do the third weighing.
If that weighing is true, you know the bad penny is the one unweighed. Now you won’t have to do the fourth, you’ve got it in by Three.
If a fourth weighing is necessary, weight just ONE of the two left, and you now can see if it comes out even to what you know one penny weighs, if it doesnt, then the last penny is the "bad" penny.
If the fourth weighing shows the penny too heavy, or two light, you also got it. No further weighings necessary.
See? I think this works.