Title
Remainder Shells
Problem Statement
Ken has between 3 and 100
shells in his collection. If he counts his shells 3 at a time, he has 2 left
over. If he counts them 4 at a time, he has 2 left over. If Ken counts them 5 at
a time, he still has 2 left over. How many shells does Ken have in his shell
collection?
Problem setup
What number is divisible
by 3 with a remainder of 2, divisible by 4 with a remainder of 2, and divisible
by 5 with a remainder of 2?
Plans to Solve/Investigate
the Problem
Since all multiples end in
0 or 5, we know that the number we are looking for ends in a 2 or 7. That
immediately limits our choices to 19 choices between 3 and 100. One solution
choice (7) exists between 3 and 10, and two choices for every ten number
increment starting with a one as the tens digit. The multiples of 3 and 4 will
also be a multiple of 5.
We shall solve this
problem as a finding the least common multiple problem. With the aid of a table
we shall list the multiples of 3, 4, and 5, and look for the first multiple they
all share. We shall start with multiples of 5 as this number will have the
smallest number of multiples less than 100.
Investigation/Exploration
of the Problem
If we add numbers of
shells starting with the first possible solution for each we shall have the set
of all possible solutions.
Number of
Shells Counting 3 at a Time 
5 
8 
11 
14 
17 
20 
23 
26 
29 
32 
35 
38 
41 
44 
47 
50 
53 
56 
59 
62 
65 
68 
71 
74 
77 
80 
83 
86 
89 
92 
95 
98 

Number of
Shells Counting 4 at a Time 
6 
10 
14 
18 
22 
26 
30 
34 
38 
42 
46 
50 
54 
58 
62 
66 
70 
74 
78 
82 
86 
90 
94 
98 

Number of
Shells Counting 5 at a Time 
7 
12 
17 
22 
27 
32 
37 
42 
47 
52 
57 
62 
67 
72 
77 
82 
87 
92 
97 

From the table we can see
that 62 is the only possible solution. The least common multiple of 5, 4, and 3
is 60. We created out table by finding multiples of 3, 4, and 5, then added two
to each solution. This lead to our solution of 62, which is also two greater
that the least common multiple.
Extension
Instead of extensions,
students might find this problem easier to solve given a little practice in
least common multiple problems. Finding the least common multiple of three
given numbers could prepare students for the list the factors method of solving
least common multiple problems presented here. The added element of “add two”
to each solution would be the additional consideration here.
Author & Contact
Pat Devane
Memorial Middle School
Conyers, GA
pjdevane@rockdale.k12.ga.us
