Divisibility Rules for 7, 11, and 12

In our previous lesson, we discussed the divisibility rules for 2, 3, 4, 5, 6, 9, and 10. In this lesson, we are going to talk about the divisibility tests for numbers 7, 11, and 12. The reason why I separated them is that the divisibility rules for 7, 11, and 12 are a little bit more advanced. However, I promise you that after learning their respective rules and applying them to some practice problems, you will realize that they are not that difficult. In fact, they are actually fun!


Divisibility Rule for 7

divisibility rule or test for 7

Rule: Take the last digit and cross it out from the original number. Then double it. Subtract it from the “new” number which is the original number excluding the last digit. If the difference is divisible by 7, then the original number must also be divisible by 7. If at first application, the result is not obviously divisible by 7, you can repeat the process as needed until you reach a two-digit number that can easily be determined if it is divisible by 7 or not. 


Example 1: True or False. The number [latex]6,895[/latex] is divisible by [latex]7[/latex].

Solution: Let’s take the last digit of [latex]6,89{\color{red}5}[/latex] which [latex]\color{red}5[/latex] then double it, thus [latex]2({\color{red}5})=10[/latex]. Now, subtract the “new” number (old number excluding the last digit) by twice the last digit, we have [latex]689-10=679[/latex]. Is [latex]679[/latex] divisible by [latex]7[/latex]? We can perform long division. But the good thing is that we can perform the process again and again until we reach a two-digit number because it is much easier to know if it’s divisible or not by [latex]7[/latex].

Let’s repeat the process one more time and see what we will get. Remember we ended up at [latex]679[/latex] from the last step. Moving on, the last digit of [latex]67{\color{red}9}[/latex] is [latex]\color{red}9[/latex]. If we double it, we get [latex]2({\color{red}9})=18[/latex]. The remaining number formed when we get rid of the last digit is [latex]67[/latex]. If we subtract [latex]67[/latex] by [latex]18[/latex], we obtain [latex]67-18=49[/latex].

Since [latex]49[/latex] is divisible by [latex]7[/latex], therefore the original number [latex]6,895[/latex] must also be divisible by [latex]7[/latex]. So, the answer is True. ✔︎


Example 2: Multiple Choice. Which number is divisible by [latex]7[/latex]?

Note: There is only one correct answer.

A) [latex]18,046[/latex]

B) [latex]11,749[/latex]

C) [latex]20,704[/latex]

D) [latex]21,011[/latex]

I understand that the procedure can be tricky at first but the more you use it, the more it becomes much easier. Below are easy-to-follow steps that I hope can help to cement in your memory.


Steps to Check for the Divisibility of 7

  • Drop the last digit of the number then double the digit that we dropped.
  • Subtract it from the new number formed by removing the last digit of the original number.
  • Repeat the process until the number is reduced to two digits.
  • If the two-digit number is divisible by [latex]7[/latex], then the original number is divisible by [latex]7[/latex]. Otherwise, it is not.

Solution: In an actual multiple questions test, you may want to randomly select an option (a letter) to solve because it is possible that you can stumble upon the correct answer right away, therefore, saving you a lot of time. But in this lesson, we will go from A to D for the sake of practice.

◉ Testing Option A: [latex]18,046[/latex]

Drop the last digit of [latex]18,046[/latex] which becomes [latex]1,804[/latex] then double the digit that we dropped, so we have [latex]2(6)=12[/latex].

Subtract the new number by the double of the last digit: [latex]1,804 – 12 = 1,792[/latex]. We have reduced the original five-digit number to a four-digit number. Remember, we want it to be reduced to a two-digit number. Let’s repeat the process.

Drop the last digit of [latex]1,792[/latex] which becomes [latex]179[/latex] then double the digit the we dropped, so we have [latex]2(2)=4[/latex].

Subtract the new number by the double of the last digit: [latex]179 – 4 = 175[/latex]. We have reduced it now to a three-digit number. Let’s do it one more time!

Drop the last digit of [latex]175[/latex] which becomes [latex]17[/latex] then double the digit that we removed, thus [latex]2(5)=10[/latex].

Subtract the new number by twice the last digit: [latex]17-10=7[/latex].

Since [latex]\color{red}7[/latex] is divisible by [latex]7[/latex], then the original number which is [latex]18,046[/latex] is also divisible by [latex]7[/latex]. So, option A is the correct answer. ✔︎

The final answer is option A.


I will leave it to you as an exercise as to why options B, C, and D are NOT divisible by [latex]7[/latex]. However, I will still provide you with a shortened solution below. I highly encourage you to perform the exercise not only for more practice but also because it is as satisfying to show that a number is not divisible by [latex]7[/latex].

You Try!

◉ Testing Option B: [latex]11,749[/latex]

Answer
  • Original number: [latex]11,749[/latex]
  • [latex]1,174-2(9)=1,174-18=1,156[/latex]
  • [latex]115-2(6)=115-12=103[/latex]
  • [latex]10-2(3)=10-6=4[/latex]

Since [latex]\color{red}4[/latex] is not divisible by [latex]7[/latex] then [latex]11,749[/latex] is also not divisible by [latex]7[/latex]. ✘

◉ Testing Option C: [latex]20,704[/latex]

Answer
  • Original number: [latex]20,704[/latex]
  • [latex]2,070-2(4)=2,070-8=2,062[/latex]
  • [latex]206-2(2)=206-4=202[/latex]
  • [latex]20-2(2)=20-4=16[/latex]

Since [latex]\color{red}16[/latex] is not divisible by [latex]7[/latex] then [latex]20,704[/latex] is also not divisible by [latex]7[/latex]. ✘

◉ Testing Option D: [latex]21,011[/latex]

Answer
  • Original number: [latex]21,011[/latex]
  • [latex]2,101-2(1)=2,101-2= 2,099[/latex]
  • [latex]209-2(9)=209-18=191[/latex]
  • [latex]19-2(1)=19-2=17[/latex]

Since [latex]\color{red}17[/latex] is not divisible by [latex]7[/latex] then the original number which is [latex]21,011[/latex] is not divisible by [latex]7[/latex] as well. ✘


Example 3: Select all that apply. Which numbers are divisible by [latex]7[/latex]?

Note: There can be more than one answer.

A) [latex]5,544[/latex]

B) [latex]3,110[/latex]

C) [latex]54,810[/latex]

D) [latex]34,125[/latex]

Solution: I am sure that at this point you have already mastered the steps on how to check if a number is divisible by [latex]7[/latex] or not. With that said, I will be using a shortened solution.

◉ Testing Option A: [latex]5,544[/latex]

We are testing if [latex]5,544[/latex] is divisible by [latex]7[/latex].

[latex]554-2(4)=554-8=546[/latex]

[latex]54-2(6)=54-12=42[/latex]

Because [latex]42[/latex] can be divided by [latex]7[/latex] then the original number [latex]5,544[/latex] is also divisible by [latex]7[/latex]. ✔︎


◉ Testing Option B: [latex]3,110[/latex]

We are checking if [latex]3,110[/latex] is divisible by [latex]7[/latex].

[latex]311-2(0)=311-0=311[/latex]

[latex]31-2(1)=31-2=29[/latex]

Since [latex]29[/latex] cannot be divided by [latex]7[/latex] then the original number [latex]3,110[/latex] is not divisible by [latex]7[/latex] either. ✘

◉ Testing Option C: [latex]54,810[/latex]

Let’s examine if [latex]54,810[/latex] is divisible by [latex]7[/latex].

[latex]5,481-2(0)=5,481-0=5,481[/latex]

[latex]548-2(1)=548-2=546[/latex]

[latex]54-2(6)=54-12=42[/latex]

The algorithm has reduced the original number into a two-digit number which is [latex]42[/latex] that is divisible by [latex]7[/latex]. It means that the original number [latex]54,810[/latex] must also be divisible by [latex]7[/latex]. ✔︎

◉ Testing Option D: [latex]34,125[/latex]

Let’s determine if [latex]34,125[/latex] is divisible by [latex]7[/latex].

[latex]3,412-2(5)=3,412-10=3,402[/latex]

[latex]340-2(2)=340-4=336[/latex]

[latex]33-2(6)=33-12=21[/latex]

We have reduced the original five-digit number into a two-digit number [latex]21[/latex] that is divisible by [latex]7[/latex]. It implies that the original number [latex]34,125[/latex] should be divisible by [latex]7[/latex] as well. ✔︎

So in summary, options A, C, and D are divisible by [latex]7[/latex].


Divisibility Rule for 11

divisibility rule or test for 11

Rule: From the left to right of a number, take the first digit, and attach an addition symbol to its left. Then subtract it by the next digit, then add the result by the third digit, and subtract again the result by the fourth digit, and so on and so forth. If the answer is divisible by 11, then the original number is divisible by 11. 

Condensed Rule: Alternately add and subtract the digits of a number from left to right. If the answer is divisible by 11, then the original number is divisible by 11.

Standard Rule: Take the alternating sum of the digits of a number. If the result is a multiple of 11, the number is divisible by 11.

NOTE: All the rules above mean the same thing. The first two rules are more instructive in nature while the last one is the rule that you may encounter in your textbook or being taught by your teacher.


Example 1: True or False. The number [latex]9,581[/latex] is divisible by [latex]11[/latex].

The rule is actually quite simple. We will add and subtract, then repeat the pattern until all the digits of the number are assigned with plus and minus symbols from left to right. After setting it up, we simplify it. If the result is a multiple of [latex]11[/latex], then the original number is also divisible by [latex]11[/latex].

Here is the set-up:

[latex]+9-5+8-1[/latex]

Step 1: [latex]+9-5=4[/latex]

[latex]4+8-1[/latex]

Step 2: [latex]4+8=12[/latex]

[latex]12-1[/latex]

Step 3: [latex]12-1=11[/latex]

[latex]11[/latex]

Since the final result is 11 and a multiple of [latex]11[/latex], then the original number which is [latex]9,581[/latex] is divisible of [latex]11[/latex]. Thus, our final answer is True. ✔︎


Example 2: Multiple Choice. Which number is divisible by [latex]11[/latex]?

Note: There is only one correct answer.

A) [latex]98,517[/latex]

B) [latex]79,829[/latex]

C) [latex]82,709[/latex]

D) [latex]50,453[/latex]


We will check the divisibility of each number from option A to option D.

◉ Checking Option A: [latex]98,517[/latex]

Let’s set it up by taking the alternating sum of the digits of the number.

[latex]9-8+5-1+7[/latex]

Then, we simplify.

[latex](9-8)+5-1+7[/latex]

[latex]1+5-1+7[/latex]

[latex](1+5)-1+7[/latex]

[latex]6-1+7[/latex]

[latex](6-1)+7[/latex]

[latex]5+7[/latex]

[latex]12[/latex]

The final result is [latex]12[/latex] which is not a multiple of [latex]11[/latex]. Therefore, the original number [latex]98,517[/latex] is not divisible by [latex]11[/latex]. ✘

◉ Checking Option B: [latex]79,829[/latex]

Set it up by writing the alternating sum of the digits.

[latex]7+9-8+2-9[/latex]

Simplify.

[latex](7+9)-8+2-9[/latex]

[latex]16-8+2-9[/latex]

[latex](16-8)+2-9[/latex]

[latex]8+2-9[/latex]

[latex](8+2)-9[/latex]

[latex]10-9[/latex]

[latex]1[/latex]

Since the final answer [latex]\large{(1)}[/latex] is not divisible by [latex]11[/latex], therefore the original number [latex]79,829[/latex] is also not divisible by [latex]11[/latex]. ✘

◉ Checking Option C: [latex]82,709[/latex]

We first construct the alternating sum of the digits of the number.

[latex]8-2+7-0+9[/latex]

Then simplify from left to right. No need to worry about the Order of Operations since we are only dealing with addition and subtraction.

[latex](8-2)+7-0+9[/latex]

[latex]6+7-0+9[/latex]

[latex](6+7)-0+9[/latex]

[latex]13-0+9[/latex]

[latex](13-0)+9[/latex]

[latex]13+9[/latex]

[latex]22[/latex]

Since the final result is [latex]22[/latex] which is a multiple of [latex]11[/latex], it implies that the original number [latex]82,709[/latex] is divisible by [latex]11[/latex]. Therefore, the final answer is C. ✔︎

☞ There is no need to check for Option D because we have already found the correct answer.

The final answer is option C.


Example 3: Which numbers are divisible by [latex]11[/latex]? Select all that apply.

Note: There can be more than one answer.

A) [latex]69,245[/latex]

B) [latex]73,186[/latex]

C) [latex]843,210[/latex]

D) [latex]918,071[/latex]

Solution:

◉ Testing Option A: [latex]69,245[/latex] if it is divisible by [latex]11[/latex]

[latex]6-9+2-4+5[/latex]

[latex]{\color{red}6-9}+2-4+5[/latex]

[latex]-3+2-4+5[/latex]

[latex]{\color{red}-3+2}-4+5[/latex]

[latex]-1-4+5[/latex]

[latex]{\color{red}-1-4}+5[/latex]

[latex]-5+5[/latex]

[latex]0[/latex]

Since [latex]0[/latex] is a multiple of [latex]11[/latex], therefore [latex]69,245[/latex] is divisible by [latex]11[/latex]. ✔︎

◉ Testing Option B: [latex]73,186[/latex] if it is divisible by [latex]11[/latex]

[latex]7-3+1-8+6[/latex]

[latex]{\color{red}7-3}+1-8+6[/latex]

[latex]4+1-8+6[/latex]

[latex]{\color{red}4+1}-8+6[/latex]

[latex]5-8+6[/latex]

[latex]{\color{red}5-8}+6[/latex]

[latex]-3+6[/latex]

[latex]3[/latex]

Since [latex]3[/latex] is not a multiple of [latex]11[/latex], thus [latex]73,186[/latex] is not divisible by [latex]11[/latex]. ✘

◉ Testing Option C: [latex]843,210[/latex] if it is divisible by [latex]11[/latex]

[latex]8-4+3-2+1-0[/latex]

[latex]{\color{red}8-4}+3-2+1-0[/latex]

[latex]4+3-2+1-0[/latex]

[latex]{\color{red}4+3}-2+1-0[/latex]

[latex]7-2+1-0[/latex]

[latex]{\color{red}7-2}+1-0[/latex]

[latex]5+1-0[/latex]

[latex]{\color{red}5+1}-0[/latex]

[latex]6-0[/latex]

[latex]6[/latex]

Since [latex]6[/latex] is not a multiple of [latex]11[/latex], hence [latex]843,210[/latex] is not divisible by [latex]11[/latex]. ✘

◉ Testing Option D: [latex]918,071[/latex] if it is divisible by [latex]11[/latex]

[latex]9-1+8-0+7-1[/latex]

[latex]{\color{red}9-1}+8-0+7-1[/latex]

[latex]8+8-0+7-1[/latex]

[latex]{\color{red}8+8}-0+7-1[/latex]

[latex]16-0+7-1[/latex]

[latex]{\color{red}16-0}+7-1[/latex]

[latex]16+7-1[/latex]

[latex]{\color{red}16+7}-1[/latex]

[latex]23-1[/latex]

[latex]22[/latex]

Since [latex]22[/latex] is a multiple of [latex]11[/latex], it implies that [latex]918,071[/latex] is divisible by [latex]11[/latex]. ✔︎

In summary, options A and D are divisible by [latex]11[/latex].


Divisibility Rule for 12

divisibility rule or test for 12

Rule: A number is divisible by [latex]12[/latex] if it is both divisible by [latex]3[/latex] and [latex]4[/latex].

  • A number is divisible by 3 if the sum of its digits is divisible by [latex]3[/latex].
  • A number is divisible by 4 if the last two digits of the number are divisible by [latex]4[/latex].

Example 1: True or False. The number [latex]7,512[/latex] is divisible by [latex]12[/latex].

Solution:

The first step is to check if it is divisible by [latex]3[/latex]. We will first add all the digits of the number of [latex]7,512[/latex].

[latex]7,512[/latex]

[latex]7+5+1+2=15[/latex]

Since [latex]15[/latex] is divisible by [latex]3[/latex], therefore [latex]7,512[/latex] is also divisible by [latex]3[/latex].

The last step is to test if the number formed by the last two digits of the original number is divisible by [latex]4[/latex], then it is divisible by [latex]4[/latex].

[latex]7,5{\color{red}12}[/latex]

Since [latex]12[/latex] is divisible by [latex]4[/latex], then [latex]7,512[/latex] is divisible by [latex]4[/latex].

Therefore, because the original number [latex]7,512[/latex] is both divisible by [latex]3[/latex] and [latex]4[/latex], then it must be divisible by [latex]12[/latex]. ✔︎


Example 2: Multiple Choice. Which number is divisible by [latex]12[/latex]?

Note: There is only one correct answer.

A) [latex]527,037[/latex]

B) [latex]981,128[/latex]

C) [latex]746,936[/latex]

D) [latex]49,9920[/latex]

Solution:

There is a faster way to test for the divisibility of [latex]12[/latex]. Remember, a number is divisible by [latex]12[/latex] if [latex]3[/latex] and [latex]4[/latex] can both divide it. Since it is much quicker to test for the divisibility of 4 than 3 because for the former you just have to look at the last two digits of the number and check if it is a multiple of [latex]4[/latex], and the latter will take slightly more time because you will have to add all the digits of the number and check if the sum is divisible by [latex]3[/latex]. Therefore, we will check first for the divisibility of [latex]4[/latex] followed by the divisibility of [latex]3[/latex]. The other way around is a little bit more time-consuming.

◉ Testing Option A: [latex]527,037[/latex] for divisibility of [latex]12[/latex]

The last two digits of [latex]527,037[/latex] is [latex]\color{red}37[/latex] is not a multiple of 4. Therefore, it is not divisible by [latex]4[/latex]. There is no need to check for the divisibility of [latex]3[/latex] since it fails on one of the two requirements. Thus, [latex]527,037[/latex] is not divisible by [latex]12[/latex]. ✘

◉ Testing Option B: [latex]981,128[/latex] for divisibility of [latex]12[/latex]

The last two digits of [latex]981,128[/latex] is [latex]\color{red}28[/latex] which is a multiple of [latex]4[/latex] that makes it divisible of [latex]4[/latex]. Now let’s check if it is divisible by [latex]3[/latex] by adding all its digits, thus 9+8+1+1+2+8=29. Since, the sum 29 is not divisible by [latex]3[/latex], then the number itself is also not divisible by [latex]3[/latex]. Because [latex]981,128[/latex] cannot be divided by both [latex]3[/latex] and [latex]4[/latex], that means the two requirements are not met, hence the original number is not divisible by [latex]12[/latex]. ✘

◉ Testing Option C: [latex]746,936[/latex] for divisibility of [latex]12[/latex]

The number [latex]\color{red}36[/latex] is the last two digits of [latex]746,936[/latex]. And it is a multiple of [latex]4[/latex] which makes the original number divisible by [latex]4[/latex]. Now for divisibility of [latex]3[/latex]. Add all the digits of [latex]746,936[/latex], we get [latex]7+4+6+9+3+6=35[/latex]. The sum of the digits is not divisible by [latex]3[/latex]. It follows that the number is also not divisible by [latex]3[/latex]. Because one of the two required conditions are not met (both are not true), then [latex]746,936[/latex] is not divisible by [latex]12[/latex]. ✘

◉ Testing Option D: [latex]49,9920[/latex] for divisibility of [latex]12[/latex]

The number [latex]20[/latex] is the last two digits of [latex]49,9920[/latex] which is clearly a multiple of [latex]4[/latex], thus makes [latex]49,9920[/latex] divisible by [latex]4[/latex]. Adding up all the digits of the number: [latex]4+9+9+9+2+0=33[/latex]. The sum [latex]33[/latex] can be divided by [latex]3[/latex] and so [latex]49,9920[/latex] is divisible by [latex]3[/latex]. Since, the original number is both divisible by [latex]3[/latex] and [latex]4[/latex], it must also be divisible by [latex]12[/latex]. ✔︎

The final answer is option D.


Example 3: Which numbers are divisible by [latex]12[/latex]? Select all that apply.

Note: There can be more than one answer.

A) [latex]344,888[/latex]

The number [latex]\color{red}88[/latex] is the last two digits of [latex]344,888[/latex] which is clearly a multiple of [latex]4[/latex], thus divisible by [latex]4[/latex].

The sum of the digits of [latex]344,888[/latex] is calculated as [latex]3+4+4+8+8+8=35[/latex]. But [latex]35[/latex] is obviously not divisible by [latex]3[/latex].

Since [latex]344,888[/latex] is only found to be divisible by [latex]4[/latex] but not by [latex]3[/latex], failing one of the two requirements implies that the original number is not divisible by [latex]12[/latex]. ✘

B) [latex]521,340[/latex]

The last two digits of [latex]521,340[/latex] formed the number [latex]\color{red}40[/latex] which is a multiple of [latex]4[/latex], thus can be divided by [latex]4[/latex].

Adding up its digits we get [latex]5+2+1+3+4+0=15[/latex]. The sum [latex]15[/latex] is divisible by [latex]3[/latex].

Since [latex]521,340[/latex] is both divisible by [latex]3[/latex] and [latex]4[/latex], then it must be divisible by [latex]12[/latex]. ✔︎

C) [latex]842,652[/latex]

The number [latex]\color{red}52[/latex] is the last two digits of the number which is clearly divisible by [latex]4[/latex].

The sum of the digits is [latex]8+4+2+6+5+2=27[/latex]. The number [latex]27[/latex] is divisible by [latex]3[/latex].

Since [latex]842,652[/latex] are both divisible by [latex]3[/latex] and [latex]4[/latex], then it should also be divisible by [latex]12[/latex]. ✔︎

D) [latex]676,968[/latex]

The last two digits [latex]\color{red}68[/latex] are divisible by [latex]4[/latex].

The sum of the digits [latex]6+7+6+9+6+8=42[/latex] is divisible by [latex]3[/latex].

Because the original number can both be divided by [latex]3[/latex] and [latex]4[/latex], then it must also be divisible by [latex]12[/latex].


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Divisibility Rules for 2, 3, 4, 5, 6, 9, and 10