Intermath | Workshop Support
 Write-up

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