a/b = 1/2; a = combos of three 1's and two 5's, for 5!/(2!3!)=10, b = combos of three 1's, one 3, and one 5, for 5!/(3!)=20 244882 skipping the middle two because the last one took me too long. the hard one... For n+1 points, you can draw n+1 lines that also go through the center of the "sphere". These points also separately make an n-dimensional shape within the "sphere". Fixing 1 point on the sphere, the other n points will satisfy the previous conditions as defined, except you only need n lines through each point and the center. With these lines fixed, all we have left to manipulate is where the n-points are located. They can either be at their original spot, or their antipode on the sphere. We know each of these lines contains the center (per the problem, must contain center), and there are two ways to place each point. Therefore, you can choose 1 out of 2 ways to place each point, or 1/2, such that the line contains the center. The total combination for all n lines together is therefore (1/2)^n, which equals 2^-n. For example, a circle is 2-dimensions so gets 3 points. One is fixed. the other two can create two lines that run through the center, each with two options to place the point on that line. So you get (1/2)^2 = 1/4. For a sphere: 4 points, 3 lines, 2 options each: (1/2)^3 = 1/8. Just wonder if that's all valid.
(0-7-10)/(1/(1-4))=51! good job! I was thinking (10-(0-7))/(1/(4-1)) when I came up with this, now that u got it you can post the next prob!
I'll throw one in. A budgie is 95% fluff by mass, obviously. The budgie's total mass is 30 grams. He gets a trim such that he is now only 90% fluff by mass. What is his mass in grams now?
Guess it's my turn to post one. Taking this from a semi-secret puzzle source of mine. Once upon a time, there was a small band of pirates. They followed a treasure map until they found the location of a treasure chest on a small island. They dug all day and all night, and behold! They uncovered a treasure chest full of gold doubloons. It was so heavy that there was quite a struggle to get it out of the hole and lug the chest back to camp. By this time, it was already late at night, so they all agreed to sleep through the night and divide the treasure in the morning. The five pirates placed their bedrolls in a circle around the treasure, which was sitting next to the fire. That way, they figured that none of them could take from the treasure while the others slept. Of course, each one figured that they were sneaky enough to do just that! At midnight, one pirate awoke, opened the chest, and counted the gold into five even piles. When he was done, there was exactly one piece remaining. To avoid arguments, he threw the spare piece into the ocean. Then, he figured that he may was well take a pile now, and then another one in the morning. So he took one pile of gold, hid it in the sand, and put the other piles back in the chest. He then went back to sleep. At one o'clock, a second pirate awoke, and he also divided the remaining gold into five equal piles. He, similarly, found one extra gold piece, and threw it into the ocean. Then, he got greedy, and decided to hide one of the piles, and put the other four back in the chest. At two, three, and four o'clock, each of the three remaining pirates sneaked a peek into the chest. Each pirate, not knowing that any of the others had touched the treasure, divided the treasure into five even piles, found one leftover piece of gold, and threw it into the sea. They then hid one of the five piles and returned the four to the chest. At dawn, each pirate arose, and after congratulating each other for being honest (remember, each pirate thinks they were the only one to take an early share of the treasure), sat down to count the gold. Lo and behold! The pile of gold still in the chest divided evenly into five parts. Each pirate takes a share, and the band splits, each one plotting to secretly return and take their hidden stash of gold. Each piece of gold is the same size and of the same value, and to simplify things, individual pieces of gold can't be split. What is the smallest number of gold pieces that could originally be in the chest for the aforementioned events to have taken place?
Well done: The first four were from easy to hard to solve for me. I The 5-digit numbers with product of the digits equal to 25, consist of two fives and three ones. There are a of these numbers, where a is the number of ways to place the three ones (because the positions of the fives is then fixed as well). The 5-digit numbers with product of the digits equal to 15, consist of one three, one five, and three ones. There are b of those numbers, where b equals the number of ways to first place the three ones, and then place the digits 3 and 5 in one of two possible ways. We see that b = 2a, which implies that a b = 1 2 . II We only consider numbers composed of the even digits 2, 4, and 8. If such a number ends with digit 4 or 8, it is a multiple of 20 plus 4 or 8, and hence divisible by 4. If such a number ends with digit 2, it is a multiple of 20 plus 2, and hence not divisible by 4. So we have to choose 2 as the last digit. Sorting the remaining digits in increasing order, we obtain 244882. III The combination of digits “2013” occurs 13 times as part of the following numbers: 2013, 12013, 22013, and 20130 to 20139. In addition, “2013” also occurs as the end of one number followed by the beginning of the next number. The different possibilities are: 2|013 does not occur, since no numbers starts with digit ‘0’. 20|13 occurs 11 times: 1320|1321 and 13020|13021 to 13920|13921. 201|3 occurs only once: 3201|3202, because no numbers larger than 30 000 were written down. It is easy to verify that “2013” does not occur as a combination of three consequtive numbers. Therefore, “2013” occurs a total of 13 + 11 + 1 = 25 times in the sequence of digits. IV Since every student answered correctly a different number of questions, at least 0 + 1 + 2 + · · · + 19 = 190 correct answers were given in total. Because every question was answered correctly at most three times, there must be at least 190 3 = 63 1 3 questions. That is, there were at least 64 questions. A situation with 64 questions is indeed possible. Let the number of correct answers given by the twenty students be 0, 1, . . . , 19. We divide the students into three groups: I consists of those students having 0, 1, 2, 3, 4, 17, 18, or 19 correct answers, II consists of those students having 5, 6, 7, 14, 15, or 16 correct answers, and III consists of those students having 8, 9, 10, 11, 12, or 13 correct answers. The total number of correct answers in each group is no more than 64. Within each group, we can therefore choose the correctly answered questions to be distinct. That way, no question is answered correctly more than three times. For the last one: indeed, that's the way to solve it. Of course, the other part is writing that down in mathematical terms and actually proving it, but, sure Also, I know 3blue1brown made a video on that problem The test I am talking about was more of an exercise-thing, the Dutch schooling system is too different for me to explain If there in total are u(n) picies currently, before, in the next step there were (5/4)u(n-1)+1, which needs to be whole for five steps. I just let my GR test all numbers, five minutes later, it stopped at 1020 for u(0), which means u(5) (which is the original amount) needs to be 3121 (I feel like I made a mistake somewhere... )
Ah, that's a well-known puzzle. I've usually heard it done with coconuts though, as then it's intuitive that you can't split them.
