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C _ C _ N _ T S Clue
HE WILL HAVE 25 COCONUTS WITH HIM AT THE END. THE TRICK IS TO REDUCE THE NUMBER OF SACKS AS YOU PASS CHECKPOINTS. THE FIRST 10 CHECKPOINTS REQUIRE 3 COCONUTS EACH, WHICH EMPTIES HIS FIRST SACK. THE NEXT 15 CHECKPOINTS REQUIRE 2 COCONUTS EACH, WHICH WILL EMPTY HIS SECOND STACK. NOW, HE IS LEFT WITH 1 SACK AND 5 MORE CHECKPOINTS. SO, THE 5 CHECKPOINTS WILL TAKE 1 COCONUT EACH. THEREFORE, HE WILL BE LEFT WITH 25 COCONUTS GOLD IS WEIGHED IN TROY OUNCES OR TROY POUNDS, WHERE THERE ARE ONLY 12 TROY OUNCES PER TROY POUND. THE MAXIMUM WEIGHT SPECIFIED ON THE BRIDGE WILL BE USING THE NORMAL IMPERIAL SYSTEM FOR MEASURING WEIGHT, WHERE THERE ARE 16 POUNDS TO AN OUNCE. BECAUSE OF THIS 36 POUNDS OF GOLD (WEIGHED USING THE TROY SYSTEM), WILL ONLY WEIGH ABOUT 29 STANDARD IMPERIAL POUNDS. THIS MEANS THAT HER TOTAL WEIGHT, INCLUDING BOTH GOLD BARS, IS LESS THAN 130 POUNDS. AN ALTERNATIVE ANSWER WOULD BE: THE GIRL WOULD SIMPLY THROW ONE OF THE GOLD BARS UP IN THE AIR WHILE HOLDING THE OTHER IN A JUGGLING MANNER. THUS, SHE WOULD NEVER BE HOLDING MORE THAN 118 POUNDS AS LONG AS SHE THREW THE OPPOSITE BAR UP BEFORE SHE CAUGHT THE OTHER WHEN THE FIRST SERVANT COMES IN, THE KING SHOULD WRITE DOWN HIS NUMBER. FOR EACH OTHER SERVANT THAT REPORTS IN, THE KING SHOULD ADD THAT SERVANT'S NUMBER TO THE CURRENT NUMBER WRITTEN ON THE PAPER, AND THEN WRITE THIS NEW NUMBER ON THE PAPER. LET X BE THE NUMBER OF THE MISSING SERVANT AND Y BE THE NUMBER THAT THE KING HAS WRITTEN. ONCE THE FINAL SERVANT HAS REPORTED IN, THE NUMBER ON THE PAPER SHOULD EQUAL: Y = (1 + 2 + 3 + ... + 99 + 100) - X (1 + 2 + 3 + ... + 99 + 100) = 5050, SO WE CAN REPHRASE THIS TO SAY THAT THE NUMBER ON THE PAPER SHOULD EQUAL: Y = 5050 - X SO TO FIGURE OUT THE MISSING SERVANT'S NUMBER, THE KING SIMPLY NEEDS TO SUBTRACT THE NUMBER WRITTEN ON HIS PAPER FROM 5050: 5050 - Y = X 9 PERFORMANCES. WE CAN HAVE A PERFORMANCE THE FIRST DAY, AND THEN EVERY THIRD DAY AFTER THAT, GIVING 8 MORE PERFORMANCES. THE TOTAL IS 1 + 8 = 9