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Eureka Math Grade 3 Module 3 Lesson 17 Answer Key

Students of Grade 3 can get a strong foundation by referring to the Eureka Math Book. It was developed by highly professional mathematics educators and the solutions prepared by them are in a concise manner for easy grasping. To achieve high scores in Grade 3, students need to solve all questions and exercises included in Eureka Grade 3 Book.

Engage NY Eureka Math 3rd Grade Module 3 Lesson 17 Answer Key

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Eureka Math Grade 3 Module 3 Lesson 17 Problem Set Answer Key

$Eureka Math Grade 3 Module 3 Lesson 17 Problem Set Answer Key 1$

a. Color all the squares with even products orange. Can an even product ever have an odd factor?

Answer: 1 x 1 = 1, 1 x 2 = 2, 1 x 3 = 3, 1 x 4 = 4, 1 x 5 = 5, 1 x 6 = 6, 1 x 7 = 8, 1 x 8 = 8, 2 x 1 = 2, 2 x 2 = 4, 2 x 3 = 6, 2 x 4 = 8, 2 x 5 = 10, 2 x 6 = 12, 2 x 7 = 14, 2 x 8 = 16, 3 x 1 = 3, 3 x 2 = 6, 3 x 3 = 9, 3 x 4 = 12, 3 x 5 = 15, 3 x 6 = 18, 3 x 7 = 21, 3 x 8 = 24, 4 x 1 = 4, 4 x 2 = 8, 4 x 3 = 12, 4 x 4 = 16, 4 x 5 = 20, 4 x 6 = 24, 4 x 7 = 28, 4 x 8 = 32, 5 x 1 = 5, 5 x 2 = 10, 5 x 3 = 15, 5 x 4 = 20, 5 x 5 = 25, 5 x 6 = 30, 5 x 7 = 35, 5 x 8 = 40, 6 x 1 = 6, 6 x 2 = 12, 6 x 3 = 18, 6 x 4 = 24, 6 x 5 = 30, 6 x 6 = 36, 6 x 7 = 42, 6 x 8 = 48, 7 x 1 = 7, 7 x 2 = 14, 7 x 3 = 21, 7 x 4 = 28, 7 x 5 = 35, 7 x 6 = 42, 7 x 7 = 49, 7 x 8 = 56, 8 x 1 = 8, 8 x 2 = 16, 8 x 3 = 24, 8 x 4 = 32, 8 x 5 = 40, 8 x 6 = 48, 8 x 7 = 56, 8 x 8 = 64.

$Eureka-Math-Grade-3-Module-3-Lesson-17-Answer Key-1$

b. Can an odd product ever have an even factor?

Answer: No, an odd product ever has an even factor.

Explanation: In the above-given question, given that, 1 x 1 = 1, 1 x 2 = 2, 1 x 3 = 3, 1 x 4 = 4, 1 x 5 = 5, 1 x 6 = 6, 1 x 7 = 8, 1 x 8 = 8, 2 x 1 = 2, 2 x 2 = 4, 2 x 3 = 6, 2 x 4 = 8, 2 x 5 = 10, 2 x 6 = 12, 2 x 7 = 14, 2 x 8 = 16, 3 x 1 = 3, 3 x 2 = 6, 3 x 3 = 9, 3 x 4 = 12, 3 x 5 = 15, 3 x 6 = 18, 3 x 7 = 21, 3 x 8 = 24, 4 x 1 = 4, 4 x 2 = 8, 4 x 3 = 12, 4 x 4 = 16, 4 x 5 = 20, 4 x 6 = 24, 4 x 7 = 28, 4 x 8 = 32, 5 x 1 = 5, 5 x 2 = 10, 5 x 3 = 15, 5 x 4 = 20, 5 x 5 = 25, 5 x 6 = 30, 5 x 7 = 35, 5 x 8 = 40, 6 x 1 = 6, 6 x 2 = 12, 6 x 3 = 18, 6 x 4 = 24, 6 x 5 = 30, 6 x 6 = 36, 6 x 7 = 42, 6 x 8 = 48, 7 x 1 = 7, 7 x 2 = 14, 7 x 3 = 21, 7 x 4 = 28, 7 x 5 = 35, 7 x 6 = 42, 7 x 7 = 49, 7 x 8 = 56, 8 x 1 = 8, 8 x 2 = 16, 8 x 3 = 24, 8 x 4 = 32, 8 x 5 = 40, 8 x 6 = 48, 8 x 7 = 56, 8 x 8 = 64. No, an odd product ever has an even factor.

c. Everyone knows that 7 × 4 = (5 × 4) + (2 × 4). Explain how this is shown in the table.

Answer: 7 x 4 = 28. 5 x 4 = 20. 2 x 4 = 8. 20 + 8 = 28.

Explanation: In the above-given question, given that, 7 x 4 = 28. 5 x 4 = 20. 2 x 4 = 8. 20 + 8 = 28.

d. Use what you know to find the product of 7 × 16 or 8 sevens + 8 sevens.

Answer: 7 x 16 = 112.

Explanation: In the above-given question, given that, 7 x 6 = 112. 8 sevens + 8 sevens. 8 x 7 + 8 x 7. 56 + 56. 112.

$Eureka Math Grade 3 Module 3 Lesson 17 Problem Set Answer Key 2$

Answer: 1 x 2 = 2, 1 x 3 = 3, 1 x 4 = 4, 1 x 5 = 5, 1 x 6 = 6, 2 x 1 = 2, 2 x 3 = 6, 2 x 4 = 8, 2 x 5 = 10, 2 x 6 = 12, 3 x 1 = 3, 3 x 2 = 6, 3 x 4 = 12, 3 x 5 = 15, 3 x 6 = 18, 4 x 1 = 4, 4 x 2 = 8, 4 x 3 = 12, 4 x 5 = 20, 4 x 6 = 24, 5 x 1 = 5, 5 x 2 = 1, 5 x 3 = 15, 5 x 4 = 20, 5 x 6 = 30, 6 x 1 = 6, 6 x 2 = 12, 6 x 3 = 18, 6 x 4 = 24, 6 x 5 = 30.

$Eureka-Math-Grade-3-Module-3-Lesson-17-Answer Key-2$

b. Draw an array to match each expression in the table below. Then, label the number of squares you added to make each new array. The first two arrays have been done for you.

Answer: 1 + 2 = 3. 3 + 6 = 9. 6 + 10 = 16. 10 + 15 = 25. 15 + 21 = 36.

$Eureka-Math-Grade-3-Module-3-Lesson-17-Answer Key-3$

c. What pattern do you notice in the number of squares that are added to each new array?

Answer: 1 + 3 = 4. 4 + 6 = 9. 6 + 10 = 16. 10 + 15 = 25. 15 + 21 = 36.

Explanation: In the above-given question, given that, 1 + 3 = 4. 4 + 6 = 9. 6 + 10 = 16. 10 + 15 = 25. 15 + 21 = 36.

d. Use the pattern you discovered in Part (b) to prove this: 9 × 9 is the sum of the first 9 odd numbers.

Answer: 9 x 9 = 81.

Explanation: In the above-given question, given that, 9 x 9 = 81. the sum of the first 9 odd numbers is 81. 9 x 9 = 81.

Eureka Math Grade 3 Module 3 Lesson 17 Exit Ticket Answer Key

Question 1. Use what you know to find the product of 8 × 12 or 6 eights + 6 eights.

Answer: 8 x 12 = 96.

Explanation: In the above-given question, given that, 8 x 12 = 96. 6 eights + 6 eights. 6 x 8 + 6 x 8. 48 + 48. 96.

Question 2. Luis says 3 × 233 = 626. Use what you learned about odd times odd to explain why Luis is wrong.

Answer: 3 x 233 = 626.

Explanation: In the above-given question, given that, 3 x 233 = 626. odd x odd = odd. but in the given question, Luis is wrong. 3 x 233 = 626. 625 + 1.

Eureka Math Grade 3 Module 3 Lesson 17 Homework Answer Key

$Eureka Math Grade 3 Module 3 Lesson 17 Homework Answer Key 4$

b. Color the rows and columns with even factors yellow.

Answer: The even factors are color with yellow.

$Eureka-Math-Grade-3-Module-3-Lesson-17-Answer Key-5$

c. What do you notice about the factors and products that are left unshaded

Answer: The factors that are left unshaded is odd numbers.

Explanation: In the above-given question, given that, the factors that are left unshaded is odd. the factors that are in diagonals is even.

d. Complete the chart by filling in each blank and writing an example for each rule.

Answer: odd times odd equals = odd. even times even equals = even. even times odd equals = even.

Explanation: In the above-given question, given that, odd times odd equals = odd. even times even equals = even. even times odd equals = even. 9 x 9 = 81. 2 x 2 = 4. 4 x 3 = 12.

e. Explain how 7 × 6 = (5 × 6) + (2 × 6) is shown in the table

Answer: 7 x 6 = 42.

Explanation: In the above-given question, given that, 7 x 6 = 42. 5 x 6 = 30. 2 x 6 = 12. ( 5 x 6 ) + (2 x 6 ). 30 + 12 = 42.

f. Use what you know to find the product of 4 × 16 or 8 fours + 8 fours.

Answer: 4 x 16 = 64.

Explanation: In the above-given question, given that, 4 x 16 = 64. 8 fours + 8 fours. 8 x 4 + 8 x 4. 32 + 32. 64.

Question 2. Today in class, we found that n × n is the sum of the first n odd numbers. Use this pattern to find the value of n for each equation below. The first is done for you. a. 1 + 3 + 5 = n × n

b. 1 + 3 + 5 + 7 = n × n

Answer: 16 = 4 x 4.

Explanation: In the above-given question, given that, 16 = 4 x 4. 1 + 3 + 5 + 7 = n x n. n = 4. 4 x 4 = 16. 1 + 3 + 5 + 7 = 16.