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Potwe 13 Ae Cp Np 25 s

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  WWW.CEMC.UWATERLOO.CA | The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem of the WeekProblem E and SolutionSound the Alarm Problem When a certain computer program runs, the user is asked to input a three-digit number of theform  abc . Each of the digits of the input must be different. The program separates the digits of the three-digit number to obtain three single distinct digits. The digits are then multipliedtogether. If the product is the same as the two-digit number formed by the tens digit and unitsdigit of the srcinal number, namely  bc , then the computer causes a siren to sound. Otherwise,the program prompts the user to try again. There are only 4 three-digit numbers that can beinput so that the program causes the siren to sound. One of the four numbers is 612. If   abc  is612, then  a × b × c  = 6 × 1 × 2 = 12 which is  bc . Find the remaining three numbers that wouldcause the siren to sound as a result of being given as input for the computer program. Solution We know that there are only 4 answers, so we could attempt a trial and error approach to findthe remaining 3 numbers. We will present a more systematic approach.First, notice that since  a, b  and  c  are digits and must, therefore, be positive integers between 0and 9. Also, since the product  a × b × c  is a two-digit number, none of   a ,  b  or  c  can equal zero.We are asked to find all three-digit numbers  abc  such that  a × b × c  = 10 b  +  c .Since  b  = 0, we can divide by  b  and the problem becomes equivalent to finding all integers a, b, c  with 1 ≤ a, b, c ≤ 9 and  a, b  and  c  distinct such that a × c  = 10 +  cb Since  a  and  c  are integers, then so is  a × c  and we must have that  cb  is an integer as well.Therefore, for each possible value of   c , we must have that  b  divides exactly into  c .We will break the problem into cases based on the value of   c . and then sub-cases based onwhat the value of   b  can be for that particular  c .Case 1:  c  = 1There are no values of   b  where  cb  is an integer and  b  =  c .Case 2:  c  = 2For  cb  to be an integer and  b  =  c , we must have  b  = 1. If   b  = 1 and  c  = 2, a × b × c  = 10 b  +  c  becomes 2 a  = 12 and so  a  = 6. Therefore, one of the three-digitnumbers is 612. This is the number that was given as an example.  WWW.CEMC.UWATERLOO.CA | The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Case 3:  c  = 3For  cb  to be an integer and  b  =  c , we must have  b  = 1. If   b  = 1 and  c  = 3, a × b × c  = 10 b  +  c  becomes 3 a  = 13 and so  a  =  133  . Since  a  is not an integer, there is nosolution in this case.Case 4:  c  = 4For  cb  to be an integer and  b  =  c , we must have  b  = 1 or  b  = 2.Case i:  b  = 1In this case,  a × b × c  = 10 b  +  c  becomes 4 a  = 14 and so  a  =  72 . Since  a  is not aninteger, there is no solution in this case.Case ii:  b  = 2In this case,  a × b × c  = 10 b + c  becomes 8 a  = 24 and so  a  = 3. Therefore, one of thethree-digit numbers is 324.Case 5:  c  = 5For  cb  to be an integer and  b  =  c , we must have  b  = 1. If   b  = 1 and  c  = 5, a × b × c  = 10 b  +  c  becomes 5 a  = 15 and so  a  = 3. Therefore, one of the three-digitnumbers is 315.Case 6:  c  = 6For  cb  to be an integer and  b  =  c , we must have  b  = 1,  b  = 2 or  b  = 3.Case i:  b  = 1In this case,  a × b × c  = 10 b  +  c  becomes 6 a  = 16 and so  a  =  83 . Since  a  is not aninteger, there is no solution in this case.Case ii:  b  = 2In this case,  a × b × c  = 10 b  +  c  becomes 12 a  = 26 and so  a  =  136  . Since  a  is not aninteger, there is no solution in this case.Case iii:  b  = 3In this case,  a × b × c  = 10 b  +  c  becomes 18 a  = 36 and so  a  = 2. Therefore, one of the three-digit numbers is 236.We can actually stop here since we have found 4 different three-digit numbers that satisfy theconditions outlined in the problem. If we had not been given the number of possible solutions,we would need to continue by checking cases when  c  = 7 , c  = 8 and  c  = 9.Therefore, the 4 three-digit numbers which cause the computer program to sound the alarmare 612, 324, 315, and 236.Note: For students who know how to program, they may wish to solve the problem in that way.
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