# What's your favorite lateral thinking puzzles? **URL:** \ **Category:** The Game Room\ **Created:** [June 5, 2010, 11:29am UTC](https://boards.straightdope.com/t/whats-your-favorite-lateral-thinking-puzzles/542042 "2010-06-05T11:29:46Z")\ **Posts on this page:** 20\ **Page:** 16
**Author:** ![otorophile](https://avatars.discourse-cdn.com/v4/letter/o/3ab097/32.png) [@otorophile](https://boards.straightdope.com/u/otorophile)\ **Post date:** [June 16, 2010, 12:42pm UTC](https://boards.straightdope.com/t/whats-your-favorite-lateral-thinking-puzzles/542042/301 "2010-06-16T12:42:42Z")
> [@Xema](#): > > For problem #1, > > Set one weight aside; divide the rest into three groups of 666. Weigh two of these groups: if unequal, you are done. If equal, you know the remaining group is either heavier or lighter. Close but not quite. 666\*3 = 1998, so you have 2 left over. ---
**Author:** ![EllisDee](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/ellisdee/32/4531_2.png) [@EllisDee](https://boards.straightdope.com/u/EllisDee)\ **Post date:** [June 16, 2010, 12:42pm UTC](https://boards.straightdope.com/t/whats-your-favorite-lateral-thinking-puzzles/542042/302 "2010-06-16T12:42:56Z")
> [@Xema](#): > > For problem #1, > > Set one weight aside; divide the rest into three groups of 666. Weigh two of these groups: if unequal, you are done. If equal, you know the remaining group is either heavier or lighter. If all three groups have exactly 333 heavy weights and 333 light weights, all three groups will weigh the same. (You meant put 2 aside, yes?) ---
**Author:** ![Nature\_s\_Call](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/nature_s_call/32/19006_2.png) [@Nature\_s\_Call](https://boards.straightdope.com/u/Nature_s_Call)\ **Post date:** [June 16, 2010, 1:26pm UTC](https://boards.straightdope.com/t/whats-your-favorite-lateral-thinking-puzzles/542042/303 "2010-06-16T13:26:40Z")
This one had to be explained to me - I just couldn’t figure it out on my own: What comes next in this series: 1, 11, 21, 1211, 111221, 11212211, 2112112221, … [spoiler]The first number is arbitrarily 1. Each subsequent number describes the digits in the last. Understood best if you say each digit out loud. That is, the second number isn’t “eleven” it’s “one one” meaning the previous number consists of how many ones? one! So next is “two one” because “11” is two ones. So the next number in the sequence is 122112213211[/spoiler] ---
**Author:** ![Peter\_Morris](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/peter_morris/32/359_2.png) [@Peter\_Morris](https://boards.straightdope.com/u/Peter_Morris)\ **Post date:** [June 16, 2010, 1:41pm UTC](https://boards.straightdope.com/t/whats-your-favorite-lateral-thinking-puzzles/542042/304 "2010-06-16T13:41:58Z")
> [@Biotop](#): > > A guy working at the corner Exxon notices a customer who stops in each morning and buys exactly 50 cents worth of gas. Sometimes (but not always) the same customer even comes back later the same day…again purchasing only and exactly 50 cents worth of gas. > > The employee can’t help but shake his head at the guy’s behavior. Finally, after watching this customer over and over, the Exxon employee approaches the chap and says, “I see you come here at least once every day. May I suggest a simple repair that would prevent you from having to come here so often. Indeed, You might not have to come back at all.” > > What’s the repair? The “gas” he bought is air, not gasoline. He can fix the problem by mending the puncture in his bicycle tire. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [June 16, 2010, 1:46pm UTC](https://boards.straightdope.com/t/whats-your-favorite-lateral-thinking-puzzles/542042/305 "2010-06-16T13:46:25Z")
Er, ignore this post. ---
**Author:** ![Peter\_Morris](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/peter_morris/32/359_2.png) [@Peter\_Morris](https://boards.straightdope.com/u/Peter_Morris)\ **Post date:** [June 16, 2010, 2:03pm UTC](https://boards.straightdope.com/t/whats-your-favorite-lateral-thinking-puzzles/542042/306 "2010-06-16T14:03:08Z")
> [@Ximenean](#): > > That reminds me of another nice one: > > Write out all the integers up to ten billion… > > Sort the list of numbers alphabetically, ignoring spaces and punctuation. What is the first odd number in the list? And what is the final number on the list? (odd or even) Zero ---
**Author:** ![Carmady](https://avatars.discourse-cdn.com/v4/letter/c/f04885/32.png) [@Carmady](https://boards.straightdope.com/u/Carmady)\ **Post date:** [June 16, 2010, 2:10pm UTC](https://boards.straightdope.com/t/whats-your-favorite-lateral-thinking-puzzles/542042/307 "2010-06-16T14:10:42Z")
The statement of problem 1 isn’t entirely clear to me, but: [spoiler]It appears from the statement that you get the actual weight difference. So put 666 vs 668. If the 668 is heaver by anything other than 198, 199, or 200 then just pull 2 off. If the 668 is heaver by 199, then switch one to the other side. If the 668 is heavier by 198 or 200, then the 666 is different than the 666 left out originally.[/spoiler] ---
**Author:** ![Peter\_Morris](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/peter_morris/32/359_2.png) [@Peter\_Morris](https://boards.straightdope.com/u/Peter_Morris)\ **Post date:** [June 16, 2010, 2:24pm UTC](https://boards.straightdope.com/t/whats-your-favorite-lateral-thinking-puzzles/542042/308 "2010-06-16T14:24:12Z")
> [@Max\_Torque](#): > > An old favorite: > > You possess a balance scale, and twelve coins. All of the coins are identical in every respect, but one. One coin is counterfeit, and it differs from the others by weight. You do not know whether the counterfeit coin is heavier or lighter, only that its weight is different from the other coins. You cannot find the counterfeit by touch..[/spoiler] A similar one. (Stolen from an episode of Columbo, where he went up against a Mensa member) You have 10 sacks of gold ingots. One of the sacks contains fake ingots, the rest are real. The real ingots weigh 10 ounces, the fake ones weigh 11 ounces. You have a set of accurate scales, which will tell you the weight of any objects placed in the pan. (Not a balance, just to be clear. No need to compare one ingot against another.) How can you find the sack of fakes in the minimum number of weightings? Assume that: - the scales are accurate to 1/10th oif an ounce - they are sturdy enough to hold any number of ingots - you can’t tell the difference in weight simply holding them in your hands - each sack contains as many ingots as you might need for the solution. [spoiler] You can do it in one weighting: Take 1 ingot from sack 1, 2 ingots from sack 2, 3 ingots from sack 3, etc. If the weight is, for instance, 7 ounces over, then you know that sack 7 is fake [/spoiler] ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [June 16, 2010, 2:50pm UTC](https://boards.straightdope.com/t/whats-your-favorite-lateral-thinking-puzzles/542042/309 "2010-06-16T14:50:32Z")
> [@otorophile](#): > > One more: You have 2n apples and 2n bananas arranged in a circle, in some arbitrary order. Can you always draw a line through the circle such that on each side of the line there are n apples and n bananas? Yes. Consider the regions along the circle inbetween fruit; there are 4n of these. Label each with (a - b)/2 where a is the number of apples and b is the number of bananas among the next 2n many fruit clockwise. Every region will be labelled with an integer, any two adjacent regions will be labelled by either the same or adjacent integers, and opposite regions (in the sense that each is the 2n-th region after the other) will have the negation of each other’s values; thus, there must be a pair of opposite regions labelled with zeros, and drawing a line through one point from each will solve the problem. ---
**Author:** ![Xema](https://avatars.discourse-cdn.com/v4/letter/x/9de053/32.png) [@Xema](https://boards.straightdope.com/u/Xema)\ **Post date:** [June 16, 2010, 3:24pm UTC](https://boards.straightdope.com/t/whats-your-favorite-lateral-thinking-puzzles/542042/310 "2010-06-16T15:24:53Z")
> [@otorophile](#): > > Close but not quite. 666\*3 = 1998, so you have 2 left over. Oops! ---
**Author:** ![Left\_Hand\_of\_Dorkness](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/left_hand_of_dorkness/32/7156_2.png) [@Left\_Hand\_of\_Dorkness](https://boards.straightdope.com/u/Left_Hand_of_Dorkness)\ **Post date:** [June 16, 2010, 4:08pm UTC](https://boards.straightdope.com/t/whats-your-favorite-lateral-thinking-puzzles/542042/311 "2010-06-16T16:08:24Z")
> [@Peter\_Morris](#): > > A similar one. (Stolen from an episode of Columbo, where he went up against a Mensa member) > > You have 10 sacks of gold ingots. One of the sacks contains fake ingots, the rest are real. The real ingots weigh 10 ounces, the fake ones weigh 11 ounces. > > You have a set of accurate scales, which will tell you the weight of any objects placed in the pan. (Not a balance, just to be clear. No need to compare one ingot against another.) > > How can you find the sack of fakes in the minimum number of weightings? > > Assume that: > > - the scales are accurate to 1/10th oif an ounce > - they are sturdy enough to hold any number of ingots > - you can’t tell the difference in weight simply holding them in your hands > - each sack contains as many ingots as you might need for the solution. > > [spoiler] You can do it in one weighting: > > Take 1 ingot from sack 1, 2 ingots from sack 2, 3 ingots from sack 3, etc. > > If the weight is, for instance, 7 ounces over, then you know that sack 7 is fake [/spoiler] Car Talk had a similar one some years back, except this time, you’re working with rolls of coins (100 coins per roll), and you have 7 rolls, and the difference between real and fake is 1 gram vs. 1.1 grams, and instead of knowing that one sack of ingots is fake, any or all of them could be fake. What’s the minimum number of weighings you need to to do figure out which, if any, rolls are fake? And then they told you: one weighing. The question is, how? ---
**Author:** ![Peter\_Morris](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/peter_morris/32/359_2.png) [@Peter\_Morris](https://boards.straightdope.com/u/Peter_Morris)\ **Post date:** [June 16, 2010, 4:25pm UTC](https://boards.straightdope.com/t/whats-your-favorite-lateral-thinking-puzzles/542042/312 "2010-06-16T16:25:24Z")
> [@Left\_Hand\_of\_Dorkness](#): > > Car Talk had a similar one some years back, except this time, you’re working with rolls of coins (100 coins per roll), and you have 7 rolls, and the difference between real and fake is 1 gram vs. 1.1 grams, and instead of knowing that one sack of ingots is fake, any or all of them could be fake. What’s the minimum number of weighings you need to to do figure out which, if any, rolls are fake? And then they told you: one weighing. The question is, how? Ah, yes. Got it. [spoiler]Use binary. Take from each roll the following r1 - 1 r2 - 2 r3 - 4 r4 - 8 r5 - 16 r6 - 32 r7 - 64 Then if the weighing is, for instance, 4.3 grams over, then that can only happen if r6, r4, r2, r1 are fake, and no others. [/spoiler] ---
**Author:** ![Xema](https://avatars.discourse-cdn.com/v4/letter/x/9de053/32.png) [@Xema](https://boards.straightdope.com/u/Xema)\ **Post date:** [June 16, 2010, 4:29pm UTC](https://boards.straightdope.com/t/whats-your-favorite-lateral-thinking-puzzles/542042/313 "2010-06-16T16:29:35Z")
> [@EllisDee](#): > > If all three groups have exactly 333 heavy weights and 333 light weights, all three groups will weigh the same. (You meant put 2 aside, yes?) No, I simply was sloppy. Here’s how it should work: 1. Divide the blocks into three groups, containing 666, 666 and 668 - call these G1, G2 and G3. 2. Weigh G1 and G2; if unequal you are done. 3. If equal, set G1 aside and move one block from G3 to G2 (which now contain 667 blocks each). 4. These two groups cannot have the same weight. Here’s why: At the start of step #3, G1 and G2 can be of equal weight only if they had the same distribution of heavy (H) and light (L) weights. So G3.H (number of heavy weights in G3) must initially be equal to 1000 - 2 \* G1.H. In other words, G3.H must be an even number. If G1.H is odd, then G2.H plus G3.H must also be odd, so no matter what block is moved, there is no way that G2 and G3 can have an equal distribution of L and H. How about if G1.H is even? Well, if G1.H = 332, then G3.H starts out as 336, so G2.H and G3.H can’t finish any closer than 333/335. Q.E.D. ---
**Author:** ![Delayed\_Reflex](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/delayed_reflex/32/3172_2.png) [@Delayed\_Reflex](https://boards.straightdope.com/u/Delayed_Reflex)\ **Post date:** [June 16, 2010, 4:44pm UTC](https://boards.straightdope.com/t/whats-your-favorite-lateral-thinking-puzzles/542042/314 "2010-06-16T16:44:11Z")
> [@Nature\_s\_Call](#): > > This one had to be explained to me - I just couldn’t figure it out on my own: > > What comes next in this series: > > 1, 11, 21, 1211, 111221, 11212211, 2112112221, … > > [spoiler]The first number is arbitrarily 1. Each subsequent number describes the digits in the last. Understood best if you say each digit out loud. That is, the second number isn’t “eleven” it’s “one one” meaning the previous number consists of how many ones? one! So next is “two one” because “11” is two ones. > > So the next number in the sequence is 122112213211[/spoiler] Shouldn’t the 6th number be 312211 if threes are allowed? Seems strange that it would be 1 one + 2 ones instead of 3 ones? If threes aren’t allowed, then I guess the next number would be 12211221122211, or alternatively, 12211221221211 😉 ---
**Author:** ![Hoopy\_Frood](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hoopy_frood/32/8452_2.png) [@Hoopy\_Frood](https://boards.straightdope.com/u/Hoopy_Frood)\ **Post date:** [June 16, 2010, 6:03pm UTC](https://boards.straightdope.com/t/whats-your-favorite-lateral-thinking-puzzles/542042/315 "2010-06-16T18:03:32Z")
> [@Delayed\_Reflex](#): > > Shouldn’t the 6th number be 312211 if threes are allowed? Seems strange that it would be 1 one + 2 ones instead of 3 ones? If threes aren’t allowed, then I guess the next number would be 12211221122211, or alternatively, 12211221221211 😉 You are correct, especially since he does go up to 3’s in his 8th number. ---
**Author:** ![Ximenean](https://avatars.discourse-cdn.com/v4/letter/x/aca169/32.png) [@Ximenean](https://boards.straightdope.com/u/Ximenean)\ **Post date:** [June 17, 2010, 3:27pm UTC](https://boards.straightdope.com/t/whats-your-favorite-lateral-thinking-puzzles/542042/316 "2010-06-17T15:27:05Z")
One of my favourite “men in hats” puzzles: Three men are in separate rooms. Someone places a white or back hat on each of their heads. They cannot see their own hats. They can observe each other on video monitors, but not communicate in any way. They are given the option of guessing the colour of their own hat, or saying nothing. They do this simultaneously. If anybody guesses incorrectly, or nobody guesses at all, they are naturally executed, that being the custom in Puzzleland. Otherwise, they are set free, or marry the king’s daughters, you know the drill. I.e. they need at least one person to guess correctly and nobody to guess incorrectly. Before this ordeal, they are allowed to agree a strategy. How can they maximise their chances of survival? ---
**Author:** ![Tom\_Scud](https://avatars.discourse-cdn.com/v4/letter/t/f08c70/32.png) [@Tom\_Scud](https://boards.straightdope.com/u/Tom_Scud)\ **Post date:** [June 17, 2010, 3:47pm UTC](https://boards.straightdope.com/t/whats-your-favorite-lateral-thinking-puzzles/542042/317 "2010-06-17T15:47:36Z")
I think I must be missing a restriction here. Can they identify who each other are on the video monitors? If so, then it’s pretty easy: [spoiler]Alan tells Bob and Carl to not guess, and tells Bob to raise his left hand if Alan is wearing a white hat, or his right hand if Alan is wearing a black hat. Then Alan guesses based on what Bob does on the video monitor. If the video monitors only show the heads (but they are identifiable) then it could be shaking/nodding the head as the sign.[/spoiler] ---
**Author:** ![Ximenean](https://avatars.discourse-cdn.com/v4/letter/x/aca169/32.png) [@Ximenean](https://boards.straightdope.com/u/Ximenean)\ **Post date:** [June 17, 2010, 3:49pm UTC](https://boards.straightdope.com/t/whats-your-favorite-lateral-thinking-puzzles/542042/318 "2010-06-17T15:49:42Z")
No communication allowed. Maybe I should have said that they are in isolated cells and simply told what colour hats the other two are wearing. ---
**Author:** ![Tom\_Scud](https://avatars.discourse-cdn.com/v4/letter/t/f08c70/32.png) [@Tom\_Scud](https://boards.straightdope.com/u/Tom_Scud)\ **Post date:** [June 17, 2010, 3:53pm UTC](https://boards.straightdope.com/t/whats-your-favorite-lateral-thinking-puzzles/542042/319 "2010-06-17T15:53:54Z")
Is anything known about the color distribution of the hats? Because if there’s nothing known, the best I can come up with is for one person to be the designated random guesser. If the hats are known to be mixed-color then the answer would be: don’t guess if you are told two different colors of hats; if you are told two same-color hats, guess the opposite for yours. ---
**Author:** ![Tom\_Scud](https://avatars.discourse-cdn.com/v4/letter/t/f08c70/32.png) [@Tom\_Scud](https://boards.straightdope.com/u/Tom_Scud)\ **Post date:** [June 17, 2010, 4:01pm UTC](https://boards.straightdope.com/t/whats-your-favorite-lateral-thinking-puzzles/542042/320 "2010-06-17T16:01:34Z")
Come to think, my spoilered strategy above would be better than designated-random-guesser even if the hats are randomly chosen and 3-black or 3-white is possible. [Previous page](https://boards.straightdope.com/t/whats-your-favorite-lateral-thinking-puzzles/542042.md?page=15) [Next page](https://boards.straightdope.com/t/whats-your-favorite-lateral-thinking-puzzles/542042.md?page=17)