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Fred W Send message Joined: 13 Jun 99 Posts: 2524 Credit: 11,954,210 RAC: 0 |
Follow-ups: At the risk of turning (sic) this into a discussion on semantics, a "turn" in that context applies only when there is more than one spider and they are taking "turns" to move. And since each corner the spider encounters involves a "turn" (change of direction) I maintain that the original question setter was guilty of poor selection of wording - which annoys the h*ll out me as there are plenty of words to choose from. F. |
Luke Send message Joined: 31 Dec 06 Posts: 2546 Credit: 817,560 RAC: 0 |
OK, here are the standings: 1. William Rothamel - 9 1/2 Points 2. Mr. Kevvy - 7 Points 3. Guido.Man - 6 Points 4. Sarge - 5 1/2 Points 5. Dominique - 4 1/2 Points 6. WinterKnight - 5 Points 7. Fred W - 4 1/2 Points 8. John McLeod VII - 1 Point 9. Labbie - 1 Point 10. Scary Capitalist - 1/2 Point 11. TBD... @ Bill - How do you suggest I get a wider audience? I have tried to attract more people unsuccessfully, maybe you could help me out here! @ Fred - I have also come to the conclusion that those two questions are poor quality wording, I have emailed the owner of the site to see if he will clear this up. Q43 - Create the number 24 using only these numbers once each: 3, 3, 7, 7. You may use only the following functions: +, -, *, /. This is not a trick question, for example the answer does not involve a number system other than base 10 and does not allow for decimal points. Thanks for all your contributions you guys! Best Regards, Luke. - Luke. |
Fred W Send message Joined: 13 Jun 99 Posts: 2524 Credit: 11,954,210 RAC: 0 |
OK, here are the standings: Oohhh! One I can understand... Q43: 7*(3 + 3/7) F. |
Luke Send message Joined: 31 Dec 06 Posts: 2546 Credit: 817,560 RAC: 0 |
OK, here are the standings: Congrats Fred! Q43: What is stated above.... Points Table: 1. William Rothamel - 9 1/2 Points 2. Mr. Kevvy - 7 Points 3. Guido.Man - 6 Points 4. Sarge - 5 1/2 Points 5. Fred W - 5 1/2 Points 6. WinterKnight - 5 Points 7. Dominique - 4 1/2 Points 8. John McLeod VII - 1 Point 9. Labbie - 1 Point 10. Scary Capitalist - 1/2 Point 11. TBD... Q44: Create the number 24 using only a 1, 3, 4, and 6. You may only use +, -, /, and *.... Best Regards, Luke. - Luke. |
William Rothamel Send message Joined: 25 Oct 06 Posts: 3756 Credit: 1,999,735 RAC: 4 |
Luke --see private messages |
Luke Send message Joined: 31 Dec 06 Posts: 2546 Credit: 817,560 RAC: 0 |
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Fred W Send message Joined: 13 Jun 99 Posts: 2524 Credit: 11,954,210 RAC: 0 |
(Bill, did you receive my PM reply?) Q44: Well, you could use 1^3 * 6 * 4 but I'm not sure if using the power is cheating. I can't think of any other way round it other than using at least one digit more than once. There again, unlike for Q43 you did not specify that the digits could be used only once each so you could use 6 * 4 / (4 * 1 - 3) F. |
Luke Send message Joined: 31 Dec 06 Posts: 2546 Credit: 817,560 RAC: 0 |
I'll have to give you it, Fred... you found a void in my question... congrats again! Fred wins Q44 Q44 Answer: 1^3 * 6 * 4 or 6 * 4 / (4 * 1 - 3) Standings: 1. William Rothamel - 9 1/2 Points 2. Mr. Kevvy - 7 Points 3. Fred W - 6 1/2 Points 4. Guido.Man - 6 Points 5. Sarge - 5 1/2 Points 6. WinterKnight - 5 Points 7. Dominique - 4 1/2 Points 8. John McLeod VII - 1 Point 9. Labbie - 1 Point 10. Scary Capitalist - 1/2 Point 11. TBD... Everyone have a go on this! Q45 : A recipe calls for 4 cups of water. You have only a 3-cup and a 5-cup container. How can you measure out four cups of water? Best Regards, Luke. - Luke. |
William Rothamel Send message Joined: 25 Oct 06 Posts: 3756 Credit: 1,999,735 RAC: 4 |
Howdy. We cooks know how to do this. 3 +5 =8 cups. So we need half. A way to get a half measurement out of a cup is to tilt it so that the liquid just touches the rim of the cup on one side and the bottom of the cup on the other. Then you will have exactly half of the volume of the cup filled with water. |
Fred W Send message Joined: 13 Jun 99 Posts: 2524 Credit: 11,954,210 RAC: 0 |
I'll have to give you it, Fred... you found a void in my question... congrats again! Fill the 5-cup container. From there fill the 3-cup container and add the remaining water from the 5-cup container to the mixture. Repeat. F. |
Luke Send message Joined: 31 Dec 06 Posts: 2546 Credit: 817,560 RAC: 0 |
That was quick you two! I'll give Bill this one, he was first on the draw by 5 minutes... good try Fred! [b]Standings:[b] 1. William Rothamel - 10 1/2 Points 2. Mr. Kevvy - 7 Points 3. Fred W - 6 1/2 Points 4. Guido.Man - 6 Points 5. Sarge - 5 1/2 Points 6. WinterKnight - 5 Points 7. Dominique - 4 1/2 Points 8. John McLeod VII - 1 Point 9. Labbie - 1 Point 10. Scary Capitalist - 1/2 Point 11. TBD... Bill breaks the 10 point mark, Fred trailing close behind... I say, whatever happened to the other 8 people that have been awarded points, this place used to be a prosperous thread... Best Regards, Luke. - Luke. |
Luke Send message Joined: 31 Dec 06 Posts: 2546 Credit: 817,560 RAC: 0 |
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W-K 666 Send message Joined: 18 May 99 Posts: 19087 Credit: 40,757,560 RAC: 67 |
cards for each floor is = (floor number * 3) -1, therefore (1 + 2 + 3 + ....... + 46 + 47) * 3 - 47 = 3447 cards using pencil and paper, no plastic brains. |
Luke Send message Joined: 31 Dec 06 Posts: 2546 Credit: 817,560 RAC: 0 |
Sorry WinterKnight, you have a incorrect answer! Please try again on the next question! Correct Answer goes to Guido.man with 3337 cards Q46 Winner: guido.man Q46 Answer: 3337 cards Q46 Solution: Note first that in all houses of cards, every level has three cards more than the level immediately above it. The number of cards for a 47-story house is therefore 2 + 2 + 3 + 2 + 3.2 + 2 + 3.3 + ... + 2 + 3.(n-1) + ... + 2 + 3.(47-1) ---------------------- 47.2 + 46.47/2 x 3 = 3337 Standings: 1. William Rothamel - 10 1/2 Points 2. Mr. Kevvy - 7 Points 3. Guido.Man - 7 Points 4. Fred W - 6 1/2 Points 5. Sarge - 5 1/2 Points 6. WinterKnight - 5 Points 7. Dominique - 4 1/2 Points 8. John McLeod VII - 1 Point 9. Labbie - 1 Point 10. Scary Capitalist - 1/2 Point 11. TBD... Q47: A shopkeeper wants to be able to dispense sugar in whole pounds ranging from one pound up to 40 pounds. He has a standard, equal-arm balance weigh scale. Being of an extremely economical outlook, he wants to use the least possible number of weights to enable him to weigh any number of pounds between 1 and 40. How many weights does he need and what are they? Best Regards to all, Luke. - Luke. |
Fred W Send message Joined: 13 Jun 99 Posts: 2524 Credit: 11,954,210 RAC: 0 |
Q47: Well, the straightforward, logical answer is 6 weights of 1, 2, 4, 8, 16lb, with the last one being anywhere between 9 and 16lb. But I am sure there is a twist that I have not spotted. F. |
William Rothamel Send message Joined: 25 Oct 06 Posts: 3756 Credit: 1,999,735 RAC: 4 |
Quick guess and musings. while I got up for a minute. You guys have an advantage since I am on GMT -6 hours and should be sleeping at this time. I remember from the past that the most compact number base was e but I don't see how that applies here. In binary he could count all the way up to 64 with 6 weights. They would be 1,2,4,8,16 and 32 lbs. This is too obvious so must not be right. maybe I should try base three but it would require two one pounders two three pounders etc--no good. Since there are usually no restrictions on specifications to these problems he could of course get by with a one pound weight and use the one pound of sugar as a weight to keep measuring in a binary fashion. and so on up to any weight that he wanted. In other words he would be making his own weights out of sugar So my answer is one 1 pound weight. |
Luke Send message Joined: 31 Dec 06 Posts: 2546 Credit: 817,560 RAC: 0 |
Both answers are wrong: It is not 1,2,4,8,16 lbs or one 1 lb.... There are actually two acceptable solutions to Q47, so please try again... Q47: A shopkeeper wants to be able to dispense sugar in whole pounds ranging from one pound up to 40 pounds. He has a standard, equal-arm balance weigh scale. Being of an extremely economical outlook, he wants to use the least possible number of weights to enable him to weigh any number of pounds between 1 and 40. How many weights does he need and what are they? Luke. - Luke. |
William Rothamel Send message Joined: 25 Oct 06 Posts: 3756 Credit: 1,999,735 RAC: 4 |
I claim that one weight is the minimum--it can be any convenient size less than or up to one pound ( 1 , 2, 4, ounces for instance) as long as you don't specify how many times he has to weigh out to get the intended amount of sugar. I can't think of a way to do it with no weights. By the way do you guys use the Metric system down there ?? Help !! Greetings DADDIO Back to sleep now for me --Maybe if i sleep on it i will see the light. |
William Rothamel Send message Joined: 25 Oct 06 Posts: 3756 Credit: 1,999,735 RAC: 4 |
Wait I see it now. He would not need any weights as long as he had the knowledge of how much one grain of sugar actually weighed and had a very precise and accurate balance. After thirty or so trials he would be up to a pound and then could converge on the answer to his required poundage. once he had done this he could keep the the one pound of sugar around to act as his standard so it would be easier for him in subsequent weighings. In fact he could keep a two and four pounder sack of sugar around as well-he would need to put the sack on the other side of the scale as well --but this would be a useful idea anyway. ANSWER --he would need no weights |
Fred W Send message Joined: 13 Jun 99 Posts: 2524 Credit: 11,954,210 RAC: 0 |
OK. I think Bill is suffering from sleep deprivation. 4 weights of 1, 3, 9, and 27lb can be combined to give any single whole number of lb's between 1 and 40. F. |
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