Solve Exercise 1 Page 98 Math 4 Textbook

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Solve exercise 1 on page 98 of Math 4 textbook regarding divisibility by 3. Which numbers are divisible by 3?

Guidance for solving exercise 1 on page 98 of Math 4 textbook

Problem:
Among the following numbers, which ones are divisible by 3?
231; 109; 1872; 8225; 92313

Solution Method:
Divisibility by 3: Similar to divisibility by 9, if the sum of the digits is divisible by 3, then the number is divisible by 3.
=> Method: Add up all the digits in the number, then divide the sum by 3. If it divides evenly, conclude that the number is divisible by 3.

Answers:
+) Number 231 has a digit sum of 2 + 3 + 1 = 6.

Since 6 is divisible by 3, number 231 is divisible by 3.

+) Number 109 has a digit sum of 1 + 0 + 9 = 10 .

Since 10 is not divisible by 3, number 109 is not divisible by 3.

+) Number 1872 has a digit sum of 1 + 8 + 7 + 2 = 18.

Since 18 is divisible by 3, number 1872 is divisible by 3.

+) Number 8225 has a digit sum of 8 + 2 + 2 + 5 = 17.

Since 17 is not divisible by 3, number 8225 is not divisible by 3.

+) Number 92,313 has a digit sum of 9 + 2 + 3 + 1 + 3 = 18.

Since 18 is divisible by 3, number 92,313 is divisible by 3.

Therefore, among the given numbers, the ones divisible by 3 are:

231; 1872; 92,313.

For solving math problems in Grade 4 on page 98, Divisibility by 3 consists of 4 exercises. Next is the explanation for solving exercise 1 on page 98 of Math 4 textbook. Mytour provides hints for answers and detailed solution methods for Solving Exercise 2 Page 98 Math 4 Textbook, Solving Exercise 3 Page 98 Math 4 Textbook, and Solving Exercise 4 Page 98 Math 4 Textbook. Students need to grasp the knowledge of this lesson well.

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