Announce to students that in this activity they will try to find all the factors of 90. Place students in pairs or small groups and allow them to work on the problem. This is a hard value! It is likely students will get stuck. It is okay to leave values blank in the first round. For instance, a group may know that 90 is a multiple of 5, but are not able to easily determine what times 5 will produce this value. MathBait™ Multiplication assumes no prior knowledge of division, and relies on students' ability to skip count. Zealous students may try skip counting by fives up to 90, however, finding other values such as what times 6 equals 90 will become tedious! Allow students to focus on determining what values they believe 90 is a multiple of, and add this to the left side - leaving the corresponding left value blank as needed.

When time has expired, ask students to share one factor pair of 90 they feel confident about. Many students will recognize 90=9×10. Announce we will use this fact to help us determine the other pairs.

Focus attention on the 10. Ask students to find a factor pair of 10. When they recognize 2×5, add this under the original expression, replacing the 10, or alternatively, display the image below to students.

Ask students if they agree with the equation; is 90 equal to 9×2×5? Circle back to the warm up if needed. Remind students we can substitute equal values. As 10=2×5 we can replace the 10 with an equal value of 2×5.

Allow students to hypothesize how this might help us to factor a number. Show them the next image and ask if they agree it is true.

Again allow students to hypothesize how this may be helpful in factoring. Explain when trying to factor 90, we might know that 90 is a multiple of 5, it is a number we say when skip counting by fives as we say every number with no ones or exactly five ones. However, knowing how many fives is challenging. Eighteen is outside of our skip counting table and while we can check by counting by fives (most students will benefit from counting by fives to convince themselves of the validity), that is a lot of fives to keep track of! We know, as long as we are substituting equal values, we haven't changed the total.

Have students work together to try to fill in the remainder of their 90 factor tree using this idea. Most students will be able to initially construct the following table. If any of these initial values are missing, provide students with the given explanation to help build problem solving and reasoning skills.

Help students as needed to find the factors listed with a ? above. These are outside our multiplication table and thus more difficult to find. However, using substitution and the commutative property we can more easily identify each.

• Factor Pair for 2: As 9×10=90 we have 9×5×2=90. We know 9×5=45, thus 45×2=90.

• Factor Pair for 3: As 9×10=90 we have 3×3×10=90. We know 3×10=30, thus 3×30=90.

• Factor Pair for 5: As 9×10=90 we have 9×2×5=90. We know 9×2=18, thus 18×5=90.

• Factor Pair for 6: As 9×10=90 we have 3×3×2×5=90. Using the commutative property we see 3×2×3×5=90 and as 3×2=6 and 3×5=15, we know 6×15=90.

Conclude by explaining we can now use one fact we know to help us find all the factors of a number! The key here is to use substitution and the commutative property. For instance, if we are unsure what times 6 makes 90, we can factor the smaller numbers in 9×10 to try to make a 6; the product of the remaining factors will produce the factor pair we are looking for!

If time permits, encourage students to find the factors of 36, using the technique above for identifying the pairs for 2 and 3, which are outside our multiplication table.