4125: Fill & Download for Free

Surprisingly, there are 40.This small program prints out them all#include <stdio.h>  int main(int argc , char **argv) {  int i1;   int i2;   int i3;   int i4;  int total = 0;  for(i1=1;i1<10;i1++) {  for(i2=0;i2<10;i2++) {  for(i3=0;i3<10;i3++) {  for(i4=0;i4<10;i4++) {  int product = i1 * i2 * i3 * i4;  if ( product == 40 ) {  total++;  printf("#%02d = %d%d%d%d\n",total,i1,i2,i3,i4);  }  }  }  }  } } #01 = 1158 #02 = 1185 #03 = 1245 #04 = 1254 #05 = 1425 #06 = 1452 #07 = 1518 #08 = 1524 #09 = 1542 #10 = 1581 #11 = 1815 #12 = 1851 #13 = 2145 #14 = 2154 #15 = 2225 #16 = 2252 #17 = 2415 #18 = 2451 #19 = 2514 #20 = 2522 #21 = 2541 #22 = 4125 #23 = 4152 #24 = 4215 #25 = 4251 #26 = 4512 #27 = 4521 #28 = 5118 #29 = 5124 #30 = 5142 #31 = 5181 #32 = 5214 #33 = 5222 #34 = 5241 #35 = 5412 #36 = 5421 #37 = 5811 #38 = 8115 #39 = 8151 #40 = 8511 

At present Rafael Nadal is no.1 in the world . Other top players are:RKPLAYERCOUNTRYMOVEMENTPOINTS1Rafael Nadal098752Roger Federer075053Andy Murray062904Alexander Zverev044005Marin Cilic041556Novak Djokovic041257Dominic Thiem039258Stan Wawrinka135409Grigor Dimitrov1345510David Goffin1305511Pablo Carreno Busta1285512Milos Raonic0269013Roberto Bautista Agut0252514Kei Nishikori0247515Kevin Anderson1247016John Isner1247017Sam Querrey2244518Jo-Wilfried Tsonga0242019Tomas Berdych1219520Jack Sock1217521Nick Kyrgios2204522Gilles Muller0187523Juan Martin del Potro1186524Lucas Pouille1182525Albert Ramos-Vinolas0175526Diego Schwartzman3172027Mischa Zverev0165228Fabio Fognini2165029Adrian Mannarino2157030David Ferrer2143531Richard Gasquet1141032Pablo Cuevas0132533Philipp Kohlschreiber0130534Feliciano Lopez0129535Andrey Rublev4124736Yuichi Sugita4120937Damir Dzumhur1119138Benoit Paire1119039Paolo Lorenzi1119040Karen Khachanov2112541Alexandr Dolgopolov3111642Gael Monfils7110543Fernando Verdasco2109044Robin Haase1108545Steve Johnson1104546Ryan Harrison699047Gilles Simon298048Aljaz Bedene296849Leonardo Mayer594450Denis Shapovalov192651Denis Istomin291552Ivo Karlovic391053Kyle Edmund690254Viktor Troicki189555Borna Coric288656Jared Donaldson088357Donald Young286558Jan-Lennard Struff1086359Nikoloz Basilashvili186260Hyeon Chung186061Yen-Hsun Lu285362Jiri Vesely082563Joao Sousa582564Daniil Medvedev081465Guido Pella080266Florian Mayer079867Horacio Zeballos079368Steve Darcis579169Peter Gojowczyk178470Federico Delbonis176071Dudi Sela171872Jordan Thompson171773Dusan Lajovic570774Evgeny Donskoy1570775Marius Copil270376Thomas Fabbiano469577Rogerio Dutra Silva266678Victor Estrella Burgos365979Filip Krajinovic1665780Cedrik-Marcel Stebe465581Norbert Gombos164382Mikhail Kukushkin063383Jeremy Chardy063084Vasek Pospisil062785Andreas Seppi161186Pierre-Hugues Herbert260787Taylor Fritz460488Janko Tipsarevic960189Laslo Djere959890Thomaz Bellucci259291Ernesto Escobedo258992Frances Tiafoe1858893Henri Laaksonen158394Malek Jaziri458095Marcos Baghdatis1058096Julien Benneteau057597Andrey Kuznetsov1056698Tennys Sandgren156699Nicolas Kicker0556100Alexander Bublik0555*Rankings provided by the ATPLast Updated: October 8, 2017

How can you get 12 with 4 rolls of a die?Step 1: Find the possible combinations of each die roll that add to 121 1 4 6 (in any order - 12 permutations)1 1 5 5 (in any order - 6 permutations)1 2 3 6 (in any order - 24 permutations)1 2 4 5 (in any order - 24 permutations)1 3 3 5 (in any order - 12 permutations)1 3 4 4 (in any order - 12 permutations)2 2 2 6 (in any order - 4 permutations)2 2 3 5 (in any order - 12 permutations)2 2 4 4 (in any order - 6 permutations)2 3 3 4 (in any order - 12 permutations) Many thanks to Sk Harrischeng as I missed this roll combination in my rush to answer the question3 3 3 3 (in any order - 1 permutation)Step 2: Count the permutations of each possible set of 4 rolls of a die (I’ve done that and plugged them into the above table - see below for the details)Solution:Total count of all permutations of all possible rolls adding to 12 is 125 out of a population of 1296 total possible roll combinations (6 x 6 x 6 x 6);probability is 125/1296 or 0.0964506 or roughly odds of 19:197 (or 5:52 is close - equivalent to 0.096153846, or about the odds of a “Royal Flush in spades” dealt in one hand (in a game with only one player at the table - the dealer) in 5 consecutive single card draws from a fresh, randomly shuffled deck of 52 playing cards)How to count the permutations of a set of four rolls of one die:Roll four dice all at the same time:(or individually one after another, it makes no difference)There are 11 possible combinations adding to 12, in various permutations.the possible permutations of 4 unique rolls is found by 4 x 3 x 2 x 1 = 24i.e. the roll 1 2 4 5 can be in any order:For the toss combination 1245 (all four rolls are unique):1245*1254*1425*1452*1524*1542*2145*2154*2415*2451*2514*2541*4125*4152*4215*4251*4512*4521*5124*5142*5214*5241*5412*5421*For the toss combination 1145 (there is one duplicate roll value):1145*1154*1415*1451*1514*1541*4115*4151*4511*5114*5141*5411*For the toss combination 2226 (there is one triplicate roll value):2226*2262*2622*6222*For the toss combination 3333 (there is one value rolled four times):3333For the toss combination 2244 (there is a pair of duplicate roll values, each unique):2244*2424*244242244242*4422*What constitutes a unique roll? That depends on the wording of the question.the roll 1145 and 1145 are similar and indistinguishable if the “ones” are reversed in order (i.e. if they appear on different dice)they are distinguishable if you use only one die and roll four times, or if you use four dice - all of differing colours*this asterisk is a legacy item and carries no meaningI should have checked here first, as it may be a trick question. Let’s start with the definition of “fair die”:... but all of the Platonic Solids can make fair dice!Fair Dice on MathIsFun (ain’t it though?)Four dice can result in totals ranging from 4 through 24. The most likely total will be a whole number midway between those limits, or 14. I see a pattern materializing…Number-of-dice / Most-likely-outcome1 / any number 1 to 6 at random2 / 73 / 10 or 114 / 145 / 17 or 186 / 21Using the example from the Fair Dice link as a guide and incorporating the permutation characteristics listed above:4 unique numbers (eg. 1234) permutes 24 ways3 unique numbers (eg. 1233) permutes 12 ways2 unique numbers (eg. 1112) permutes 4 waysa pair of 2 unique (eg. 1122) numbers permutes 6 ways1 unique number (eg. 1111) permutes only 1 wayTotal roll tally - probability (roll1, roll2, roll3 …; permute1, permute2, …)where probability = total permutations / total possible sample size4 - 1/1296 (1111, there is only 1 way to get 4)5 - 4/1296 (1112, permuted 4 ways)6 - 10/1296 (1113, 1122, permuted 4 and 6 ways respectively)7 - 16/1296 (1114, 1123; 4, 12)8 - 22/1296 (1115, 1124, 1133; 4, 12, 6)9 - 28/1296 (1116, 1125, 1134; 4, 12, 12)10 - 30/1296 (1126, 1135, 1144; 12, 12, 6)11 - 80/1296 (1136, 1145, 1226, 1235, 1244, 1334; 12, 12, 6, 24, 12, 12)12 - 125/1296 (1146, 1155, 1236, 1245, 1335, 1344, 2226, 2235, 2244, 2334, 3333; 12, 6, 24, 24, 12, 12, 4, 12, 6, 12, 1)13 - 154/1296 (?, ?…; ?, ?…)14 - 170/1296 (?, ?…; ?, ?…) ←- this one should have the greatest probability, is my guess15 - 154/1296 (?, ?…; ?, ?…)16 - 125/1296 (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?; 12, 6, 24, 24, 12, 12, 4, 12, 6, 12, 1) ←- this one should be 125/1296 (from symmetry)17 - 80/1296 (?, ?, ?, ?, ?, ?; 12, 12, 6, 24, 12, 12) ←- this one should be 80/1296 (from symmetry)18 - 30/1296 (?, ?, ?; 12, 12, 6) ←- this one should be 30/1296 (from symmetry)19 - 28/1296 (?, ?, ?; 4, 12, 12) ←- this one should be 28/1296 (from symmetry)20 - 22/1296 (?, ?, ?; 4, 12, 6) ←- this one should be 22/1296 (from symmetry)21 - 16/1296 (?, ?; 4, 12) ←- this one should be 16/1296 (from symmetry)22 - 10/1296 (?, ? ; 4, 6) ←- this one should be 10/1296 (from symmetry)23 - 4/1296 (5666; 4)24 - 1/1296 (6666; 1)And as the author of the Fair Die site mentioned… I did not reduce my fractions. In this case I did not even express the probabilities as a decimal number from 0 to 1.Work in progress beyond this ^point…Further note: by calculating the odds of every possible outcome, we can add the totals to confirm my logic. The total of all possible random 4-die toss probabilities should equal 1.i.e.1 + 4 + 10 + 16 + 22 + 28 + 30 + 80 + 125 + (?) + (?) + (?) + 125 + 80 + 30 + 28 + 22 + 16 + 10 + 4 + 1 = 12961 + 4 + 10 + 16 + 22 + 28 + 30 + 80 + 125 + (154) + (170) + (154) + 125 + 80 + 30 + 28 + 22 + 16 + 10 + 4 + 1 = 1110 (not quite, need another 186 permutations)We need only calculate the probabilities of rolling the totals 13 and 14 to complete the field, as symmetry proves the rest of the scale.i.e.p(14) + 2 x p(13) = 664p(170) + 2 x p(154) = 478 (still missing 664 - 478 = 186 permutations)To that end, I propose:A) an analysis of the number of combinations possible to bring 4 rolls of a die to 13:1156 (1) permutes 12 times1246 (2) 241255 (3) 121336 (4) 121345 (5) 241426*1435*1444 (6) 6 was it 6 or 41516*1525*1534*1615*1624*1633*2146*2155 (7) 122236 (8) 122245 (9) 122326*2335 (10) 122344 (11) 122416*2425*2434*2515*2524*2533*2614*2623*2634*3136*3145*3226*3235*3316*3325*3334 (12) 43416*3425*3433*3514*3523*3613*3622*4126*41 done*42 done*43 done*4414*4423*I think that covers it… I just need to remove the duplicates (*) from the aboveMaybe there are only 12 combinationsI found a total of 12 + 24 + 12 + 12 + 24 + 6 + 12 + 12 + 12 + 12 + 12 + 4 = 154 permutations possible for 13*duplicates included only to keep the sequence in my head… to be removed from the final count laterB) an analysis of the number of combinations possible to bring 4 rolls of a die to 14:1166 (1) permutes 6 ways1256 (2) 241346 (3) 241355 (4) 121436*1445 (5) 121454*1526*1535*1544*1616*1625*1634*2156*2246 (6) 122255 (7) 62336 (8) 122345 (9) 242426*2435*2444 (10) 42516*2525*2534*2615*2624*2633*3335 (11) 43344 (12) 63434*3533*3614*4442*4514*4523*4613*I see a pattern… that seems to be itnow remove the duplicate (*) items from the listI found 12 combinations totaling 14I found a total of 6 + 24 + 24 + 12 + 12 + 12 + 6 + 12 + 24 + 4 + 4 + 6 = 170 permutations possible for 14Next step is to apply the permutation rule as before…. I will go back over both lists now with those numbers….Still need another 186 permutations - but I am out of time for now…. more later