Can you solve the giant iron riddle? - Alex Gendler
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Date: 2020-08-22
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Comments and reviews: 10
Amyprost
So, I made a bit of experimentation for the last riddle. If I'm not wrong, if an electrical component is well fabricated, it only is current that makes it work. If you want to turn on an LED with 220 volts, but it only runs with 2 amps, you must have the apropriate resistor for the LED to work correctly. The only way you can make the iron work by placing only 4 batteries, even though it has 6 available slots, i that you connect all batteries in a parallel circuit. The total voltage across the entire circuit, assuming that all batteries produce the same voltage, is the voltage of one battery. If all batteries produce n volts, the net voltage is also of n volts. Furthermore, if only one battery has a different voltage, the circuit will overheat and one of the main batteries, first or last/ top or bottom, will explode do to the amount of current. Actually, for no battery to explode, all batteries should have the same voltage. Also, the first and last batteries are the main ones because of the arrangement: the current through any other battery is considerably smaller than the main ones, therefore, being almost a null contribution for the functioning of the circuit.
This makes the iron only needing to have one battery to work, instead of 4.
reply
So, I made a bit of experimentation for the last riddle. If I'm not wrong, if an electrical component is well fabricated, it only is current that makes it work. If you want to turn on an LED with 220 volts, but it only runs with 2 amps, you must have the apropriate resistor for the LED to work correctly. The only way you can make the iron work by placing only 4 batteries, even though it has 6 available slots, i that you connect all batteries in a parallel circuit. The total voltage across the entire circuit, assuming that all batteries produce the same voltage, is the voltage of one battery. If all batteries produce n volts, the net voltage is also of n volts. Furthermore, if only one battery has a different voltage, the circuit will overheat and one of the main batteries, first or last/ top or bottom, will explode do to the amount of current. Actually, for no battery to explode, all batteries should have the same voltage. Also, the first and last batteries are the main ones because of the arrangement: the current through any other battery is considerably smaller than the main ones, therefore, being almost a null contribution for the functioning of the circuit.
This makes the iron only needing to have one battery to work, instead of 4.
reply
Chtchav
Other way to resolve the riddle:
Test batteries AB => iron doesn't start
Test batteries BC => iron doesn't start
Therefore, in the 4 batteries DEFG that are left, in the worst case there are at least 2 good batteries. It can be 3 or 4 but let's assume it's 2.
Put aside battery G and use 3 test to test DEF. => iron doesn't start
Therefore G has to be a good battery, and there are exactly 2 bad batteries amongst DEF (maximum that is possible. If there had been 1 bad battery or 0 bad battery amongst DEF, a combination would have started the iron.
You have 2 more tries and you isolated G that is a good battery.
Take batteries A and B. You know at least one of them is a good one. It cant be the 2 of them is a bad battery otherwise the DEF test would have started the iron.
Test A en G => doesn't start.
Then you know B and G are good batteries!
reply
Other way to resolve the riddle:
Test batteries AB => iron doesn't start
Test batteries BC => iron doesn't start
Therefore, in the 4 batteries DEFG that are left, in the worst case there are at least 2 good batteries. It can be 3 or 4 but let's assume it's 2.
Put aside battery G and use 3 test to test DEF. => iron doesn't start
Therefore G has to be a good battery, and there are exactly 2 bad batteries amongst DEF (maximum that is possible. If there had been 1 bad battery or 0 bad battery amongst DEF, a combination would have started the iron.
You have 2 more tries and you isolated G that is a good battery.
Take batteries A and B. You know at least one of them is a good one. It cant be the 2 of them is a bad battery otherwise the DEF test would have started the iron.
Test A en G => doesn't start.
Then you know B and G are good batteries!
reply
arnav
there is another easy solution try pairs of 2, one. of them must work and if don't it means that in each there is one good and bad battery, cause if both bad then one has to be both good, then just pick any of 2 you are sure that there are 2 working so take 1 of the battery and try 2 from anther, if even this not works then you have done only 6 attempts and. now we are 100 percent sure that these left 2 will work, in this the work case will take 7 tries but almost 80 percent of cases will be solved in 4 or less
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there is another easy solution try pairs of 2, one. of them must work and if don't it means that in each there is one good and bad battery, cause if both bad then one has to be both good, then just pick any of 2 you are sure that there are 2 working so take 1 of the battery and try 2 from anther, if even this not works then you have done only 6 attempts and. now we are 100 percent sure that these left 2 will work, in this the work case will take 7 tries but almost 80 percent of cases will be solved in 4 or less
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Rine
We can try with four batterys
Lets number it: -1, 2, 3, 4
First u tried 1 and 2 battery (by let's assume worst )iron is not working now insert battery 3 and 4 iron is still not working
So now we have 5 try left first lets insert battery 1 and 3 if iron is still not working lets insert battery 1 and 4 if its not working lets insert battery 2, 3 and 2, 4
Now we have 1 try left
If none of the combination work then other four batterys are working.
reply
We can try with four batterys
Lets number it: -1, 2, 3, 4
First u tried 1 and 2 battery (by let's assume worst )iron is not working now insert battery 3 and 4 iron is still not working
So now we have 5 try left first lets insert battery 1 and 3 if iron is still not working lets insert battery 1 and 4 if its not working lets insert battery 2, 3 and 2, 4
Now we have 1 try left
If none of the combination work then other four batterys are working.
reply
Ngan
answer to bonus riddle is 4 tries:
ABCDEF
you try AB, CD, since there are 4 working ones within the 6, you will be able to find the 2 working ones within 2 tries.
so say EF has the good 2, then AB and CD has one good battery each, so then you try AC and AD to find the remaining good 2.
If AC and AD works then you found working batteries
If AC and AD doesn't work then it must be B & C.
reply
answer to bonus riddle is 4 tries:
ABCDEF
you try AB, CD, since there are 4 working ones within the 6, you will be able to find the 2 working ones within 2 tries.
so say EF has the good 2, then AB and CD has one good battery each, so then you try AC and AD to find the remaining good 2.
If AC and AD works then you found working batteries
If AC and AD doesn't work then it must be B & C.
reply
vikrant
I tried it in different way,
There are only two possible ways to arrange 4 good and 4 bad batteries
1. First at least on good pair is in sequence
That's AB BC CD DE EF FG one of them should be good
2. If not then alternate sequence should be good and that would be
AC CE EG
but we don't need to try all if all earlier tries are made from #1 and non work then AC has to be good
reply
I tried it in different way,
There are only two possible ways to arrange 4 good and 4 bad batteries
1. First at least on good pair is in sequence
That's AB BC CD DE EF FG one of them should be good
2. If not then alternate sequence should be good and that would be
AC CE EG
but we don't need to try all if all earlier tries are made from #1 and non work then AC has to be good
reply
ukasz
There's actually a more straightforward solution which guarantees finding a working pair in 7 tries or less. Divide into 4 pairs, test each pair. Worst case scenario is each pair has a working and a not working battery, which means you can take two pairs and test one of the batteries with the remaining three (maximum of 3 tries.
reply
There's actually a more straightforward solution which guarantees finding a working pair in 7 tries or less. Divide into 4 pairs, test each pair. Worst case scenario is each pair has a working and a not working battery, which means you can take two pairs and test one of the batteries with the remaining three (maximum of 3 tries.
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Noah
Here is my answer for six tries:
Pair them all up, try all four pairs. If none work, all four pairs have one working and one not. Take any two pairs and switch one battery from each. One pair is guaranteed to work while the other is guaranteed to not. Try both pairs. Worst case scenario is 6 tries.
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Here is my answer for six tries:
Pair them all up, try all four pairs. If none work, all four pairs have one working and one not. Take any two pairs and switch one battery from each. One pair is guaranteed to work while the other is guaranteed to not. Try both pairs. Worst case scenario is 6 tries.
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Nick
Lick each battery, duh. Worst case, you lick 6 batteries and the last two give you a shock. Then your 7th try can be placing the two good batteries into the iron. If you think the shock will be too much for you, test them on the baby.
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Lick each battery, duh. Worst case, you lick 6 batteries and the last two give you a shock. Then your 7th try can be placing the two good batteries into the iron. If you think the shock will be too much for you, test them on the baby.
reply
Chandir
The second riddle is exactly the same riddle as the first one, isn't it? You pretend to test the ones that are not in the iron and you'll need 7 or less tries with the same technique.
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The second riddle is exactly the same riddle as the first one, isn't it? You pretend to test the ones that are not in the iron and you'll need 7 or less tries with the same technique.
reply
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