In this activity, we will reinforce the ideas developed in Ninety to help students easily determine the factor pair with values outside the multiplication table.

Begin by asking students the value of 8×5. If students have completed the previous units in MathBait™ Multiplication, they will be able to identify the product of 40. Explain this product is within our multiplication table so we are familiar with it, when values are outside the table they can be more tricky. We are going to learn how to make the tricky problems easier by forcing them back into our table.

Tell students to suppose they did not know what 8×5 was, but they are really good at counting by tens and so they know 8×10=80. Ask how they might use the fact 8×10=80 to find 8×5. Allow students to share their thoughts and ideas. Highlight the previous activity, as 8×10=8×2×5. How might we use this fact to help us find 8×5?

There are in fact multiple ways to go about this!

Method 1: Halving

Students are not yet familiar with division, so this is a fun way to get us thinking about halves. Since 8×2×5=80 we know 16×5=80 as well. But our goal here was to find 8×5. Looking closely, we can see 8×5 is hidden in our expression 8×2×5=80. This tells us 8×10 is double (or two times) 8×5. Ask students how we could split 80 into two equal groups. In other words, what number could we add to itself to find 8? Students should be able to determine 4+4=8 and thus 4 tens + 4 tens = 8 tens or alternatively 40+40=80. If students are not flexible enough with addition, this might be challenging. 

Method 2: Building 40

If your students are not yet ready to consider halving, stick to the methods shown in the activity Ninety. An alternative method to determine 8×5 is to break up the 8. There are many ways to achieve this. Encourage students to find and share different methods. One way is to write 8 as 2×4, which makes 8×5=2×4×5. For here, students can notice this expression is equivalent to 10×4 or to 2×20, both easier facts to find. This is a good opportunity to get them thinking about halving by noticing 4 is half of 8, or two fours make 8. This view is further helping students in factoring as they are now thinking about how to break a number into 2 times something.

For the remainder of the activity, have students work independently or in pairs on additional problems.

• 12×5

• 13×5

• 14×5

• 15×5

• 16×5

• 17×5

Every problem except 13 and 17 can be broken down into a known fact. Here are some possible solutions:

• 12×5=2×6×5=2×30=60

• 12×5=3×4×5=3×20=60

• 14×5=7×2×5=7×10=70

• 15×5=3×5×5=3×25=75 (this is particularly good for students with a solid understanding of money)

• 16×5=2×8×5=2×40=80

• 16×5=8×2×5=8×10=80

Students will likely have a hard time with 13 and 17. In the next activities we will re-introduce prime numbers so this is a great time to discuss why these values are so hard to find! We introduced prime numbers in MathBait™ Multiplication Part 2. Remind students as needed that a prime number is a tower height that can only be built with blocks which are 1-unit tall or, if we have a large block, can be built with one large block equal to the number. The values 13 and 17 are prime. We can only write them as 1×13 or 1×17, making our strategy not very effective.

Ask students if we can think of another way to determine 13×5 or 17×5, or any tricky value when dealing with a prime number. One way which we have explored previously in MathBait™ Multiplication is using the distributive property. We can find,

13×5 = (10+3)×5 = (10×5)+(3×5) = 50+15=65.

Another method, not previous discussed, is to use the common theme of manipulation. We previously found 12×5=60. We have learned this simply means we counted by 5's twelve times. Thus, thirteen times is simply one more 5, and so 13×5=60+5=65. Encourage students to "try on" both methods of working with tricky primes to find 17×5 and discuss what they liked and disliked about the process.