Date: 12/18/95 at 22:13:59 From: Anonymous Subject: Geometry puzzles I am puzzled. I would very much appreciate if you could solve these two puzzles I cannot get. In these puzzles you are to determine a 3-digit number (no digit repeated) by making "educated guesses." After each guess, you will be given a clue about your guess. The clues: bagels: no digit is correct pico: one digit is correct but in the wrong position fermi: one digit is correct and in the correct position. In each of the problems a number of guesses have been made with the clue for each guess shown to the right. Form the given set of clues, and determine the 3-digit number. If there is more than one solution, find them all. 1. 123 pico, 456 pico, 789 pico, 941 bagels, 375 pico, 638 pico 2. 198 pico fermi, 765 bagels, 432 pico, 129 pico fermi Thanks :)
Date: 3/2/96 at 11:18:58 From: Doctor Jodi Subject: Re: Geometry puzzles Hi! This is a wonderful puzzle you sent us. First of all, if order doesn't matter, and you're choosing 3 distinct numbers from 1 to 9, there are 9X8X7 = 504 possible answers, since for the first number you can pick any number from 1-9 (9 choices), the second time you can choose any number from 1-9 EXCEPT the one you picked (8 Choices), and the third time you can choose any number from 1-9 except the TWO you just picked (7 choices). Fortunately, the hints pare those answers down pretty quickly! Next, you need to know that for any group of 3 numbers, there are 6 different ways to write that group. Say we have the group 1, 2, 3. We can write that as 123 132 231 213 321 312 Now, let's look at the hints. For each bagel, we know we can't use any of those numbers. For each pico, we MUST use one of those numbers - but NOT in the place given. And for each fermi, we MUST use one of those numbers in EXACTLY the place given. So, for 941, a bagel, we can't use any of those numbers. That leaves us with 2, 3, 5, 6, 7, 8 as choices.... But that's still a lot. Let's see what else we can find out about the solution. In each of the picos, (123, 456, 789, 375, 638), we know that NONE of the numbers are in the right place, but that we MUST use exactly one number from each group. Since 3 appears, first, second, and third, we can also eliminate it; since it CAN'T be in any of those places, it can't be in our final number at all. Since we just eliminated 1, 3, 4, and 9, and since we must use exactly one number from each of the picos, we know that we must use 2 5 OR 6 7 OR 8 7 OR 5 6 OR 8 We can rewrite this if 5, not 6, not 7 if 6, not 5, not 8 if 7, not 5, not 8 if 8, not 6, not 7 Since we must use exactly 3 numbers from 2, 5, 6, 7, 8, and since we must use 2 to satisfy the pico 123, we can now see that if we use 5, we must use 8 (the only choice left, after using 2 and eliminating 6 and 7), giving us the group 2, 5, 8 if we use 6, we must use 7 (similarly, the only choice left, after using 2 and eliminating 5 and 8), giving us the group 2, 6, 7 We still need to check to see if all of the 12 answers we now have (as we said before, 6 from each group of three numbers) will work. We still haven't used one fact: none of the numbers in our pico's is in the right place. Remember, those were 123, 456, 789, 375, and 638. The number 2--can't be middle, so it must be first or last 5--can't be middle or last, so it must be first 6--can't be last or first, so it must be middle 7--can't be first or middle, so it must be last 8--can't be middle or last, so it must be first Now let's combine this with our groups of numbers--2, 5, 8, and 2, 6, 7 - to figure out which orders of these numbers will work. For 2, 5, 8, 8 must be first, 5 must be first, and 2 must be first or last. We can't make 5 and 8 BOTH first at the same time, so NONE of these solutions will work. What about 2, 6, and 7? 2 must be first or last, 6 must be middle, and 7 must be last. This gives us the number 267, which is the solution to the first puzzle. Think you can figure out the second one on your own? Good luck, and write us back if you need more help. --Doctor Jodi, The Math Forum
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