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Elementary Mathematics

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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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