Solving Exercise 2 Page 99 Math 4 Textbook

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Ngày cập nhật gần nhất: 15/4/2026

Solving exercise 2 on page 99 Math 5 Textbook in Focus Practice, Among the numbers 57 234; 64 620; 5270; 77 285

Guidance for solving exercise 2 page 99 Math 4 Textbook

Problem:
Among the numbers 57 234; 64 620; 5270; 77 285
a) Which numbers are divisible by 2 and 5 ?
b) Which numbers are divisible by 3 and 2 ?
c) Which numbers are divisible by 2, 3, 5, and 9 ?

Solution Method:
- A number divisible by both 2 and 5 has its last digit as 0
- A number divisible by both 3 and 5 satisfies both conditions: Its digit sum is divisible by 3, and its last digit is either 0 or 5
- A number divisible by 2, 3, 5, and 9 must meet the following conditions simultaneously: The last digit is 0, the sum of its digits is divisible by both 3 and 9.

Answer:

a) Numbers ending in 0 are divisible by both 2 and 5.

Hence, the numbers divisible by both 2 and 5 are: 64 620; 5270.

b) Numbers divisible by 2 are 57 234; 64 620; 5270.

Number 57 234 has a digit sum of 21; number 64 620 has a digit sum of 18; number 5270 has a digit sum of 14.

Therefore, numbers divisible by 3 are: 57 234; 64 620.

Hence, numbers divisible by both 3 and 2 are: 57 234; 64 620.

c) We can select from the numbers in part b to find numbers divisible by 5 and 9.

Among the two numbers 57 234; 64 620, the number divisible by 5 is 64 620.

Furthermore, number 64 620 has a digit sum of 18, so it is divisible by 9.

Hence, numbers divisible by both 2, 3, 5, and 9 are: 64620.

Section solving math problems for grade 4 page 99 comprises 4 exercises, after solving exercise 2 page 99 Math 4 Textbook, students are invited to check out the hints and methods for Solving Exercise 1 Page 99 Math 4 Textbook, Solving Exercise 3 Page 99 Math 4 Textbook, Solving Exercise 4 Page 99 Math 4 Textbook and Solving Exercise 5 Page 99 Math 4 Textbook, to excel in grade 4 mathematics.

Frequently Asked Questions

Which numbers among the given are divisible by both 2 and 5?

The numbers that are divisible by both 2 and 5 are those ending in 0. In this case, 64,620 and 5,270 meet this criterion, as their last digits are 0.

How can students determine if numbers are divisible by 3 and 2?

To determine if numbers are divisible by both 3 and 2, first, check if they are even for divisibility by 2. Then, calculate the sum of their digits; if the sum is divisible by 3, the number qualifies.

What are the conditions for divisibility by 2, 3, 5, and 9 simultaneously?

A number must satisfy several conditions to be divisible by 2, 3, 5, and 9: it must end in 0, and the sum of its digits must be divisible by both 3 and 9 to qualify.

What should students do after solving exercise 2 on page 99?

After solving exercise 2 on page 99, students should refer to hints and methods for other exercises, including exercises 1, 3, 4, and 5, to enhance their understanding of grade 4 mathematics.

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