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=> Find the Latest Grade 6 Math Problem-Solving Materials Here: Grade 6 Math Solutions
- Grade 6 Workbook, Page 33, Connecting Knowledge Series - Exercise 8: Divisibility Relationship and Properties
- Grade 6 Workbook, Page 46, Kite Series - Exercise 11: Decomposing a Number into Prime Factors
- Grade 6 Workbook, Pages 38-39, Horizon of Creativity Series - Exercise 12: Common Factors. Greatest Common Factor
Guide to Grade 6 Workbook, Pages 38-39 Problem Solving (Brief)
1. Grade 6 Math Problem 1, Pages 38-39, Exercise 91
Among the following numbers, which are divisible by 2, which are divisible by 5 ?
652; 850; 1546; 785; 6321
Solution:
Numbers divisible by 2 and 5 are:
652 : divisible by 2
850 : divisible by 2 and 5
1546 : divisible by 2
785 : divisible by 5
6321 : not divisible by 2 or 5
2. Grade 6 Math Problem 1, Pages 38-39, Exercise 92
Given the numbers 2141; 1345; 4620; 234. Among these numbers:
a) Which numbers are divisible by 2 but not by 5 ?
b) Which numbers are divisible by 5 but not by 2 ?
c) Which numbers are divisible by both 2 and 5 ?
Solution:
a) Numbers divisible by 2 but not by 5 are: 234.
b) Numbers divisible by 5 but not by 2 are: 1345.
c) Numbers divisible by both 2 and 5 are: 4620.
3. Solving Grade 6 Math Problems, Chapter 1, Pages 38-39, Exercise 93
Find out if the following sums (differences) are divisible by 2 or 5:
a) 136+420;
b) 625−450
c) 1.2.3.4.5.6+42
d) 1.2.3.4.5.6−35
Solution:
a) 136 is divisible by 2, not by 5 and 420 is divisible by 2 and 5 => 136 + 420 is divisible by 2 and not by 5
b) 625 is not divisible by 2, divisible by 5 and 450 is divisible by 2 and 5 => 625 - 450 is divisible by 5 and not by 5
c) 1.2.3.4.5.6 = 720 is divisible by 2, by 5 and 42 is divisible by 2, not by 5 => 1.2.3.4.5.6+42 is divisible by 2 and not by 5
d) 1.2.3.4.5.6 = 720 is divisible by 2, by 5 and 35 is not divisible by 2, divisible by 5 => 1.2.3.4.5.6−35 is not divisible by 2, divisible by 5
4. Solving Grade 6 Math Problems, Chapter 1, Pages 38-39, Exercise 94
Without performing division, find the remainder when each of the following numbers is divided by 2, by 5
813; 264; 736; 6547.
Solution:
- Divided by 2
813 : 2 = (812 + 1) : 2 => Remainder 1
264 is divisible by 2 => No remainder
736 is divisible by 2 => No remainder
6547 : 2 = (6546 + 1) : 2 => Remainder 1
- Divided by 5
813 : 5 = (810 + 3) : 5 => Remainder 3
264 : 5 = (260 + 4) : 5 => Remainder 4
736 : 5 = (735 + 1) : 5 => Remainder 1
6547 : 5 = (6545 + 2) : 5 => Remainder 2
5. Solving Grade 6 Math Problems, Chapter 1, Pages 38-39, Exercise 95
Fill in the digit for the * to get the number 54* satisfying the conditions
a) Divisible by 2
b) Divisible by 5
Solution:
a) To be divisible by two, the digit must be even, with the units digit being even. Therefore, 0, 2, 4, 6, 8 are suitable digits to satisfy the condition.
b) To be divisible by 5, the digit must end with 0 or 5. Therefore, 0, 5 are suitable digits to satisfy the condition.
6. Solve Exercise 96 on page 39 of Grade 6 Mathematics Textbook, Chapter 1
Fill in the digit for the * to get the number *85 satisfying the conditions
a) Divisible by 2
b) Divisible by 5
Solution:
a) To be divisible by two, the digit must be even, with the units digit being even. Since the last digit of *85 is 5, it cannot be divisible by 2.
b) To be divisible by 5, the digit must end with 0 or 5. Therefore, any digit can be filled in for *85 to be divisible by 5.
Solving Grade 6 math problem set 1 Signs of division by 2, by 5 page 38, 39 exercise 97
Using the digits 4, 0, 5, form three-digit natural numbers with different digits satisfying the conditions:
a) The number is divisible by 2
b) The number is divisible by 5
Solution:
a) For divisibility by 2, the last digit must be even => we have the numbers: 504, 540, 450
b) For divisibility by 5, the last digit must be 0 or 5 => we have the numbers: 450, 405, 540
Solving Grade 6 math problem set 1 page 39 exercise 98
Mark 'X' in the appropriate box in the following sentences:
Solution:
Solving math problem 6 set 1 page 38, 39 exercise 99
Find the natural number with two digits, the digits being the same, knowing that when divided by 2 and divided by 5, the remainder is 3.
Solution:
Solve problem 100 on page 39 of Math 6 workbook set 1
The first car was invented in a year outside
Solution:
Through the section Solving Grade 6 math problem set 1 page 38, 39 Signs of division by 2, by 5, arranged by difficulty level from basic to advanced with exercises: Finding numbers divisible by 2, by 5 among given numbers, finding sums/differences divisible by 2 or 5, finding remainders when dividing given numbers by 2 and 5 without actually performing division,... somewhat helped students understand more about the signs of division by 2, division by 5, signs of division by both 2 and 5. Besides, these exercises also helped students develop flexibility in completing various types of math problems and reinforce the theoretical part already learned to lay the foundation for advanced mathematical knowledge in the future.
Next, students can refer to the math problems solved in Grade 6 workbook pages 35, 36 or preview the solutions in Grade 6 workbook pages 41, 42 to improve their understanding of Grade 6 Math.
