Pattern Problems

We have begun to explore patterns.  Patterns are found everywhere, especially in math.  Today’s question:

Looking at these three figures…

Answer this question…

(Following the pattern of the first three figures, can you find the number of red, blue, and the total number of tiles used, in the fifth and tenth figure?)

Some students solved the problem by drawing each of the figures, from 1 to 10:

Some students were able to find a pattern in the length of one ‘leg’ of the figures, allowing them to move straight to the fifth and tenth figures:

Some students used a chart to look at the various numbers of tiles:

Consolidation:  We discussed how using a chart makes the number patterns jump out.  Patterns appear in a variety of ways, including: horizontally, vertically, and diagonally.  Using charts, therefore, is an effective way of making predictions. 

Patters were described as… Starting at a number, and then… (increasing, decreasing, or repeating)

An interesting question was, can we use a math algorithm to predict a pattern on a chart.  (To be explored…)

Today’s question:

Some students solved the problem by drawing all the possible solutions from 1 table to 15 tables:

Some students were able to apply the pattern they saw, and only drew the figures that had 5 and 15 tables:

Some students were able to see a mathematical pattern that they followed (repetitive addition) to find the answer:

Some students were able to use a table to solve the problem:

Consolidation: We discussed why using a table to find the answer to pattern problems is a more efficient way to solve problems.  The reasons that were brought up were written on the above solution.

We also noted that using a table allows us to see, and explain, number patterns, which often lead to algorithms that can help us predict answers without having to fill in the table all the way to the missing number.  Some of the groups even began to find a way of using algorithms in their answers:

In this case, realizing that one table holds 6 people (therefore 6 X 15 = 90) and then subtracting the people who will not be able to sit in between each table.  There are 14 sides where no one can sit if the tables are pushed together, where normally 2 people would sit – one at each table.  Therefore you subtract 28 from 90 and you get 62.  Very creative thinking!

This group realized that 15 tables have 2 people on one side (15 x 2) and 2 people on the other side (15 x 2) and then 1 person on each end (1 x 2).  Altogether that makes 62 people.  Very creative problem solving!


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