A number a is divisible through the number b if a \div b has actually a remainder the zero (0). Because that example, 15 divided by 3 is specifically 5 which indicates that that remainder is zero. Us then say the 15 is divisible by 3.

You are watching: Numbers divisible by 2 and 3

In our other lesson, we disputed the divisibility rules for 7, 11, and also 12. This time, we will certainly cover the divisibility rule or exam for**2**, **3**, **4**, **5**, **6**, **9**, and also **10**. Believe me, friend will have the ability to learn them very quickly due to the fact that you might not understand that you currently have a straightforward and intuitive knowledge of it. For instance, it is obvious that all also numbers space divisible by 2. That is pretty much the divisibility dominion for **2**. The score of this divisibility rule lesson is to formalize what you already know.

Divisibility rules aid us to determine if a number is divisible by one more without going with the actual department process such together the long division method. If the number in question are numerically little enough, we might not have to use the rules to test for divisibility. However, fornumbers whose worths are big enough, we want to have some rule to serve as “shortcuts” to aid us figure out if castle are undoubtedly divisible by every other.

**A number is divisible by2if the critical digit the the number is 0, 2, 4, 6, or 8.**

Example 1: Is the number 246 divisible by 2?

Solution: since the last digit that the number 246 end in 6, that means it is divisible through 2.

Example 2: i m sorry of the numbers 100, 514, 309, and 768 are divisible by 2?

Solution: If we study all 4 numbers, only the number 309 doesn’t end with 0, 2, 4, 6, or 8. We have the right to conclude the all the numbers above except 309 are divisible through 2.

**A number is divisible by 3 if the sum of the number of the number is divisible by 3.**

Example 1: Is the number 111 divisible by 3?

Solution:** **Let’s include the digits of the number 111. We have actually 1 + 1 + 1 = 3. Due to the fact that the amount of the number is divisible by the 3, thus the number 111 is likewise divisible by 3.

Example 2: Which one of the 2 numbers 522 and also 713 is divisible by 3?

Solution: The sum of the digits of 522 (5+2+2=9) is 9 i m sorry is divisible by 3. That provides 522 divisible through 3. However, the number 713 has actually 11 as the sum of its number which is plainly not divisible by 3 thus 713 is no divisible through 3. Therefore, only 522 is divisible by 3.

**A number is divisible by 4 if the last 2 digits of the number are divisible by 4.**

Example 1: What is the just number in the set below is divisible through 4?

945, 736, 118, 429

Solution:** **Observe the last two digits the the 4 numbers in the set. An alert that 736 is the only number within the last 2 digits (36) is divisible by 4. We deserve to conclude that 736 is the only number in the set that is divisible by 4.

Example 2: True or False. The number 5,554 is divisible by 4.

Solution:** **The last 2 digits of the number 5,554 is 54 i beg your pardon is not divisible by 4. That method the offered number is no divisible through 4 therefore the prize is **false**.

**A number is divisible by 5 if the critical digit the the number is 0 or 5.**

Example 1: many Choice. I beg your pardon number is divisible through 5.

*A)* 68

*B)* 71

*C)* 20

*D)* 44

Solution: In order because that a number to it is in divisible through 5, the critical digit that the number should be either 0 or 5. Going end the choices, just the number 20 is divisible by 5 therefore the prize is choice** C**.

Example 2: choose all the numbers that are divisible by 5.

*A)* 27

B) 105

*C)*556

*D)* 343

E) 600

Solution: Both 105 and also 600 are divisible through 5 because they either end in 0 or 5. Thus, alternatives **B** and **E** space the correct answers.

**A number is divisible by 6 if the number is divisible by both 2 and also 3.**

Example 1: Is the number 255 divisible by 6?

Solution: For the number 255 to be divisible by 6, it must divisible by 2 and 3. Let’s check first if the is divisible by 2. Note that 255 is no an also number (any number ending in 0, 2, 4, 6, or 8) which makes it no divisible 2. There’s no need to check further. We deserve to now conclude the this is not divisible through 6. The prize is **NO**.

Example 2: Is the number 4,608 divisible through 6?

Solution: A number is an also number so it is divisible through 2. Now inspect if that is divisible by 3. Let’s do that by adding all the number of 4,608 i m sorry is 4 + 6+ 0 + 8 = 18. Obviously, the sum of the number is divisible through 3 since 18 ÷ 3 = 6. Since the number 4,608 is both divisible by 2 and also 3 climate it must likewise be divisible by 6. The answer is **YES**.

**A number is divisible by 9 if the amount of the digits is divisible by 9.**

Example 1: Is the number 1,764 divisible by 9?

Solution: For a number to it is in divisible by 9, the amount of its number must also be divisible through 9. For the number 1,764 we get 1 + 7 + 6 + 4 = 18. Due to the fact that the sum of the digits is 18 and is divisible by 9 therefore 1,764 should be divisible through 9.

Example 2: select all the numbers that room divisible through 9.

*A)* 7,065

*B)* 3,512

*C)* 8,874

*D)* 22,778

*E)* 48,069

Solution:** **Let’s include the digits of every number and check if its sum is divisible through 9.

**NOT**divisible by 9.For 8,874, 8 + 8 + 7 + 4 = 27 which is divisible by 9.For 22,778, 2 + 2 + 7 + 7 + 8 = 26 i beg your pardon is

**NOT**divisible by 9.For 48,069, 4 + 8 + 0 + 6 + 9 = 27 i m sorry is divisible by 9.

Therefore, choices **A**, **C**, and **E** room divisible by 9.

**A number is divisible by 10 if the critical digit that the number is 0.See more: In What Types Of Organisms Does Meiosis Occur, Meiosis Flashcards**

The number 20, 40, 50, 170, and also 990 are all divisible by 10 due to the fact that their critical digit is zero, 0. Top top the various other hand, 21, 34, 127, and 468 are not divisible through 10 due to the fact that they don’t finish with zero.

**You might additionally be interested in:**

Divisibility Rules because that 7, 11, and 12

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