You’re blindfolded and given a pile of coins. You’re told that 99 of those coins are heads up, and the rest are tails. You may flip as many coins as you wish, as many times as you want, but the only way you’ll be allowed to remove the blindfold is if you manage to divide the coins into 2 piles (not necessarily equal piles)in which both piles have an equal amount of heads (up). How can you accomplish that?
You’re blindfolded and given a pile of coins. You’re told that 99 of those coins are heads up, and the rest are tails. You may flip as many coins as you wish, as many times as you want, but the only way you’ll be allowed to remove the blindfold is if you manage to divide the coins into 2 piles (not necessarily equal piles)in which both piles have an equal amount of heads (up). How can you accomplish that?
I'm not sure I understand the premises of the riddle...
So you have more than 99 coins spread out in front of you, 99 are heads, the others are tail... You don't know which are heads and which are tails and you cannot find out because you are blingsfolded... (I suppose the idea is not to feel them up and identify if they are heads or tails)... You could count the total number of coins, but the number of coins is not relevant...
All that is relevant is to split the coins in a way that there are an equal number of heads in both piles...
Oh, isn't that a bit similar to the locker riddle Malki posted a while ago?
I'm not sure I understand the premises of the riddle...
So you have more than 99 coins spread out in front of you, 99 are heads, the others are tail... You don't know which are heads and which are tails and you cannot find out because you are blingsfolded... (I suppose the idea is not to feel them up and identify if they are heads or tails)... You could count the total number of coins, but the number of coins is not relevant...
All that is relevant is to split the coins in a way that there are an equal number of heads in both piles...
Oh, isn't that a bit similar to the locker riddle Malki posted a while ago?
You perfectly understood it.
I don’t think it’s in any way related to the locker riddle.
This is how they answered the circuit breaker riddle , very simplified.
Hidden:
For the ease of keeping track of things, put a piece of masking tape on each circuit breaker and on each light. On the first trip to the basement, flip 50 circuit breakers to off, mark these circuit breakers with a “0,” and mark the 50 circuit breakers that are on with a “1.” Accordingly, as you roam around the house to tally the lights, mark the 50 lights that are off with a “0” and mark the other 50 lights with a “1.”
On your second trip to the basement, keep off half of the circuit breakers that are marked with a “0,” turn off half of the circuit breakers that are marked with a “1,” and mark all of these circuit breakers with a second number of “0.” Flip on all other circuit breakers if they’re not already on, and mark their second number as “1.” Now go around the house, and again mark the lights that are off with a “0” and those lights that are on with a “1.”
At the end of this step, all of your circuit breakers and lights should be marked with either “00,” “11,” “10,” or “01.” In fact, you’ve completely separated the matching problem into four different groups of 25—I.e., all lights must be matched to a circuit breaker with their same two-digit code.
You’ll continue this process: In the third trip, flip half (or actually, 13 since 25 is an odd number) of all of the circuit breakers in each group ( “00,” “11,” “10,” and “01”) to off, and mark them with an additional “0.” Mark the 12 “on” circuit breakers in each group with a “1.” Go around the house, and once again mark all lights that are off with a “0” and all lights that are on with a “1.”
Now, you’ll have created eight different groups of either 12 or 13 lights and circuit breakers: “000,” “100,” “011,” “111,” “010,” “110,” “001,” and “111.” The lights still must be matched to a circuit breaker with the same three-digit string.
After your fourth trip, you’ll have subdivided the groups into 16 groups with either six or seven lights and circuit breakers in each group. After the fifth trip, you’ll have 32 groups with three or four lights and circuit breakers in each. And after the sixth trip, you’ll have either one or two lights or circuit breakers in each group.
For those groups that each have one light and one circuit breaker, you’ve successfully mapped those circuit breakers to their lights! For the rest, it takes only one more trip—the seventh trip—to finally map them to their respective lights.
If you’re familiar with binary numbers, you’ll recognize that there are exactly 128 numbers that use seven digits, 0000000 to 1111111, which may help to explain why this strategy works so well: Each circuit breaker and light ends up with a unique “code” that maps to a specific binary number.
Using this strategy, eight trips would allow you to map up to 256 circuit breakers, nine trips would get you to 512, and 10 trips would map up to 1,024!
This is how they answered the circuit breaker riddle , very simplified.
Hidden:
For the ease of keeping track of things, put a piece of masking tape on each circuit breaker and on each light. On the first trip to the basement, flip 50 circuit breakers to off, mark these circuit breakers with a “0,” and mark the 50 circuit breakers that are on with a “1.” Accordingly, as you roam around the house to tally the lights, mark the 50 lights that are off with a “0” and mark the other 50 lights with a “1.”
On your second trip to the basement, keep off half of the circuit breakers that are marked with a “0,” turn off half of the circuit breakers that are marked with a “1,” and mark all of these circuit breakers with a second number of “0.” Flip on all other circuit breakers if they’re not already on, and mark their second number as “1.” Now go around the house, and again mark the lights that are off with a “0” and those lights that are on with a “1.”
At the end of this step, all of your circuit breakers and lights should be marked with either “00,” “11,” “10,” or “01.” In fact, you’ve completely separated the matching problem into four different groups of 25—I.e., all lights must be matched to a circuit breaker with their same two-digit code.
You’ll continue this process: In the third trip, flip half (or actually, 13 since 25 is an odd number) of all of the circuit breakers in each group ( “00,” “11,” “10,” and “01”) to off, and mark them with an additional “0.” Mark the 12 “on” circuit breakers in each group with a “1.” Go around the house, and once again mark all lights that are off with a “0” and all lights that are on with a “1.”
Now, you’ll have created eight different groups of either 12 or 13 lights and circuit breakers: “000,” “100,” “011,” “111,” “010,” “110,” “001,” and “111.” The lights still must be matched to a circuit breaker with the same three-digit string.
After your fourth trip, you’ll have subdivided the groups into 16 groups with either six or seven lights and circuit breakers in each group. After the fifth trip, you’ll have 32 groups with three or four lights and circuit breakers in each. And after the sixth trip, you’ll have either one or two lights or circuit breakers in each group.
For those groups that each have one light and one circuit breaker, you’ve successfully mapped those circuit breakers to their lights! For the rest, it takes only one more trip—the seventh trip—to finally map them to their respective lights.
If you’re familiar with binary numbers, you’ll recognize that there are exactly 128 numbers that use seven digits, 0000000 to 1111111, which may help to explain why this strategy works so well: Each circuit breaker and light ends up with a unique “code” that maps to a specific binary number.
Using this strategy, eight trips would allow you to map up to 256 circuit breakers, nine trips would get you to 512, and 10 trips would map up to 1,024!
7, 6 will give u most of the answers but when you keep sub dividing since here are odd numbers there will be some groups that need an additional trip for the answer.
You’re blindfolded and given a pile of coins. You’re told that 99 of those coins are heads up, and the rest are tails. You may flip as many coins as you wish, as many times as you want, but the only way you’ll be allowed to remove the blindfold is if you manage to divide the coins into 2 piles (not necessarily equal piles)in which both piles have an equal amount of heads (up). How can you accomplish that?
Hidden:
You count them to see the difference till 99. Whatever the difference is that’s how many you flip. Let’s say there are 150. You flip 51 and divide them in equal piles.
If it’s an odd number you flip one less than the difference and divide them. The pile with the extra one will be one with a tail anyway.
In both cases you first divide the first 99 before you start flipping. The first flip goes to the lesser pile.
You count them to see the difference till 99. Whatever the difference is that’s how many you flip. Let’s say there are 150. You flip 51 and divide them in equal piles.
If it’s an odd number you flip one less than the difference and divide them. The pile with the extra one will be one with a tail anyway.
In both cases you first divide the first 99 before you start flipping. The first flip goes to the lesser pile.
Hidden:
Trying to get it, so you first divide 99 in 2 piles, flip the rest and continue dividing.....it’s not adding up in my head.....
This is how they answered the circuit breaker riddle , very simplified.
Hidden:
For the ease of keeping track of things, put a piece of masking tape on each circuit breaker and on each light. On the first trip to the basement, flip 50 circuit breakers to off, mark these circuit breakers with a “0,” and mark the 50 circuit breakers that are on with a “1.” Accordingly, as you roam around the house to tally the lights, mark the 50 lights that are off with a “0” and mark the other 50 lights with a “1.”
On your second trip to the basement, keep off half of the circuit breakers that are marked with a “0,” turn off half of the circuit breakers that are marked with a “1,” and mark all of these circuit breakers with a second number of “0.” Flip on all other circuit breakers if they’re not already on, and mark their second number as “1.” Now go around the house, and again mark the lights that are off with a “0” and those lights that are on with a “1.”
At the end of this step, all of your circuit breakers and lights should be marked with either “00,” “11,” “10,” or “01.” In fact, you’ve completely separated the matching problem into four different groups of 25—I.e., all lights must be matched to a circuit breaker with their same two-digit code.
You’ll continue this process: In the third trip, flip half (or actually, 13 since 25 is an odd number) of all of the circuit breakers in each group ( “00,” “11,” “10,” and “01”) to off, and mark them with an additional “0.” Mark the 12 “on” circuit breakers in each group with a “1.” Go around the house, and once again mark all lights that are off with a “0” and all lights that are on with a “1.”
Now, you’ll have created eight different groups of either 12 or 13 lights and circuit breakers: “000,” “100,” “011,” “111,” “010,” “110,” “001,” and “111.” The lights still must be matched to a circuit breaker with the same three-digit string.
After your fourth trip, you’ll have subdivided the groups into 16 groups with either six or seven lights and circuit breakers in each group. After the fifth trip, you’ll have 32 groups with three or four lights and circuit breakers in each. And after the sixth trip, you’ll have either one or two lights or circuit breakers in each group.
For those groups that each have one light and one circuit breaker, you’ve successfully mapped those circuit breakers to their lights! For the rest, it takes only one more trip—the seventh trip—to finally map them to their respective lights.
If you’re familiar with binary numbers, you’ll recognize that there are exactly 128 numbers that use seven digits, 0000000 to 1111111, which may help to explain why this strategy works so well: Each circuit breaker and light ends up with a unique “code” that maps to a specific binary number.
Using this strategy, eight trips would allow you to map up to 256 circuit breakers, nine trips would get you to 512, and 10 trips would map up to 1,024!
I prefer my solution... it's more or less the same, but more orderly...
You’re blindfolded and given a pile of coins. You’re told that 99 of those coins are heads up, and the rest are tails. You may flip as many coins as you wish, as many times as you want, but the only way you’ll be allowed to remove the blindfold is if you manage to divide the coins into 2 piles (not necessarily equal piles)in which both piles have an equal amount of heads (up). How can you accomplish that?
Hidden:
Put exactly 99 coins in pile A, and all the rest it pile B.
Now flip all coins from side A.
And Viola!
Since there were exactly 99 heads, then whatever amount in pile A that was NOT heads, will reflect the amount of heads in pile B. For instance, say there were 50 heads in pile A, then 49 heads must be in pile B, which is the same amount of tails in pile A. Now that all 49 tails from A became heads, they equal amt in pile B.
Put exactly 99 coins in pile A, and all the rest it pile B.
Now flip all coins from side A.
And Viola!
Since there were exactly 99 heads, then whatever amount in pile A that was NOT heads, will reflect the amount of heads in pile B. For instance, say there were 50 heads in pile A, then 49 heads must be in pile B, which is the same amount of tails in pile A. Now that all 49 tails from A became heads, they equal amt in pile B.
Slowly slowly,
Hidden:
What if there are 110 coins total. Can you walk me through exactly how you end up with 2 piles of exactly the same amount of heads? And what if there are 500?
