# Potwe 13 Ae Cp Np 25 s

Description
Elementary Mathematics
Categories
Published

View again

All materials on our website are shared by users. If you have any questions about copyright issues, please report us to resolve them. We are always happy to assist you.
Related Documents
Share
Transcript
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.

Jul 23, 2017

#### BIL Cases 1

Jul 23, 2017
Search
Similar documents

View more...
Tags

Related Search