## Math 6 Chapter 1 Lesson 10: Divisibility of a sum

## 1. Summary of theory Tóm

### 1.1. Recall the divisibility relationship.

We know: A natural number a is divisible by a non-zero natural number b if there is a natural number k such that a = b . k.

The notation: a divisible by b is a \( \vdots \) b

a is not divisible by b is a \(\not \vdots \) b

### 1.2. Feature 1.

If a \( \vdots \) m and b \( \vdots \) m then (a + b) \( \vdots \) m :

a \( \vdots \) m and b \( \vdots \) m (a + b) \( \vdots \) m

Or you can write: (a + b) \( \vdots \) m or a + b \( \vdots \) m.

**Attention :**

a) Property 1 is also true for a difference (a \(\geq\) b) :

a \( \vdots \) m and b \( \vdots \) m ⇒ (a – b) \( \vdots \) m.

b) Property 1 is also true for a sum of many terms:

a \( \vdots \) m and b \( \vdots \) m and c \( \vdots \) m ⇒ (a + b + c) \( \vdots \) m.

If all terms of a sum are divisible by the same number, then the sum is divisible by that number.

a \( \vdots \) m and b \( \vdots \) m and c \( \vdots \) m ⇒ (a + b + c) \( \vdots \) m.

### 1.3. Feature 2.

If a \(\not \vdots \) m and b \( \vdots \) m then (a + b) \(\not \vdots \) m :

a \(\not \vdots \) m and b \( \vdots \) m ⇒ (a + b) \(\not \vdots \) m.

**Attention :**

a) Property 2 is also true for a difference (a > b):

a \(\not \vdots \) m and b \( \vdots \) m ⇒ (a – b) \(\not \vdots \) m.

b) Property 1 is also true for a sum with many terms, where only one term is not divisible by m, the remaining terms are all divisible by m:

a \(\not \vdots \) m, b \( \vdots \) m and c \( \vdots \) m⇒ (a + b + c) \(\not \vdots \) m.

If only one term of the sum is not divisible by a number and all other terms are divisible by that number, then the sum is not divisible by that number.

a \(\not \vdots \) m, b \( \vdots \) m and c \( \vdots \) m⇒ (a + b + c) \(\not \vdots \) m.

## 2. Illustrated exercise

**Question 1:** Without adding up, consider whether the sum 27+81+63 is divisible by 3?

**Solution guide:**

We see \(\)27 \( \vdots \) 3 ; 81 \( \vdots \) 3 ; 63 \( \vdots \)3 so 27+81+63 can be inferred \( \vdots \) 3

**Verse 2:** Without adding the sum, consider whether the sum of 21+49+32 is divisible by 7?

**Solution guide:**

We see 21 \( \vdots \) 7 ; 49 \( \vdots \) 7 ; 32 \(\not \vdots \) 7 so 21+49+32 \(\not \vdots \) 7 can be inferred.

**Question 3: **Without calculating the difference, consider whether the difference 42 – 18 is divisible by 6?

**Solution guide:**

We see 42\( \vdots \) 6 ; 18 \( \vdots \) 6 so the difference can be derived from 42 – 18 \( \vdots \) 6.

## 3. Practice

### 3.1. Essay exercises

**Question 1: **Without adding the sum, consider whether the sum 42+123+579 is divisible by 3?

**Verse 2:** Without the sum, consider whether the sum 24+32+56 is divisible by 8?

**Question 3:** Without calculating the signal, consider whether the difference 63 – 49 is divisible by 7?

### 3.2. Multiple choice exercises Bài

**Question 1: **Given the sum \(A = 14 + 16 + 18 + 20.\) Based on the divisibility of a sum, will A be divisible by?

A. 2

B. 5

C. 7

D. 8

**Verse 2: **Considering the sum \(30 + 15 + 90\) will be divisible by ?

A. 2

B. 6

C. 4

D. 5

**Question 3: **Which of the following is divisible by 7?

A. \(49 – 35 – 7\)

B. \(50 – 36 -8\)

C. \(80 – 17 – 14\)

D. \(79 – 19 – 15\)

**Question 4: **Considering the sum \(81 + 270 + 72\) will be divisible by?

A. 7

B. 8

C. 9

D. 10

**Question 5: **\(127 . 6 + 36\) will be divisible by?

A. 6

B. 5

C. 9

D. 8

## 4. Conclusion

Through this lecture, the divisibility of a sum, you need to complete some of the objectives given by the lesson, such as:

- Master the divisibility properties of a sum.
- Do related math problems.

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