Author	Thread: Can you find the solution to this problem? (Math)
T0TH3M4X View Profile History	Can you find the solution to this problem? (Math) Posted : 6 Dec, 2019 11:48 PM Here is a math problem. See if you can solve it. - You have a balance scale, and 8 balls that look alike and weigh the same, except for one ball that is slightly heavier than the others. What's the least amount of times you can use the scale in order to guarantee that you have found the heavier ball? How did you come up with that number? (This question is straightforward and there is no hidden trickery) Post Reply
LittleDavid View Profile History	Can you find the solution to this problem? (Math) Posted : 7 Dec, 2019 05:00 AM Question this quote “...and 8 balls that look alike and weigh the same, except for one ball that is slightly heavier than the others.” ? I’m asking because at first you say “8 weigh the same, except for one...” which is mathematically imprecise and creates some confusion an apparent mathematical disjunction between “8 that weigh the same” and “except for 1” which seems to introduce a separate or additional ball or 1 ball which weighs the same except it weighs more? I understand there’s no trick, so I’m just checking to verify that I understand the mathematical riddle you’ve introduced Are you saying there are, “ 8 balls that look alike and 7 of these 8 balls weigh the same but 1 of the 8 is slightly heavier than the other 7” ? If this is the case, you would use the weight balancer only 2 times to determine the heaviest ball. However, this assumes you’re using an equal arm balance as opposed to a gravity scale or a digital scale which would require a minimum of 4 weighings—l think Post Reply
T0TH3M4X View Profile History	Can you find the solution to this problem? (Math) Posted : 7 Dec, 2019 07:10 AM 7 balls same weight 1 ball slightly heavier 1 balance scale All eight balls look the same. Post Reply
View Profile History	Can you find the solution to this problem? (Math) Posted : 25 Mar, 2020 02:45 PM How do you figure only 2? Post Reply
T0TH3M4X View Profile History	Can you find the solution to this problem? (Math) Posted : 28 Mar, 2020 07:12 PM The answer is 2. Weigh 3 to 3, if they balance even, then it's in the remaining two. If one drops lower, it's in that pile. Regardless of the outcome, you can measure 1 to 1 in the next round . If the balance even, it's the ball you didn't weight. If one drops lower on the scale, then that would be the ball. Post Reply
View Profile History	Can you find the solution to this problem? (Math) Posted : 28 Mar, 2020 09:12 PM That makes sense. Post Reply