Divisibility Rules

When learning about multiples and divisors, there are several rules of divisibility that a student may encounter. Below, we list some famous rules of divisibility:

When dividing by...

2: The last digit is even.

3: The sum of digits is a multiple of 3.

4: The last 2 numbers are a multiple of 4.

5: The last digit is either 0 or 5.

6: Multiple of 2 and 3

8: The last 3 digits are a multiple of 8.

9: The sum of digits is a multiple of 9.

10: The last digit is 0.

11: The alternating sum of digits is a multiple of 11.

We will show the proofs of rules of 8 and 11. The rest follow in a similar manner.

Proof: When a number is divisible by 8, the last 3 digits are a multiple of 8.

Let the number be N=1000M+100a+10b+c N= 1000M + 100a + 10b + c, where a,b,c a, b, c are digits and M M is a non-negative integer.

Clearly, 1000M 1000M is a multiple of 8, since 82353 8 \mid 2^3 \cdot 5^3. Hence, N N is a multiple of 8 if and only if 100a+10b+c 100a + 10b + c is a multiple of 8.

Proof: When a number is divisible by 11, the alternating sum of digits is a multiple of 11.

From Factorization , we know that

10n(1)n=(10(1))(10n1+10n2(1)++(1)n1) 10^n - (-1)^n = (10 - (-1) ) ( 10^{n-1} + 10^{n-2}(-1) + \ldots + (-1)^{n-1} )

is always a multiple of 11.

Hence, if N=10kak+10k1ak1++10a1+a0 N = 10^k a_k + 10^{k-1} a_{k-1} + \ldots + 10 a_1 + a_0, then

N=[(10k(1)k)ak+(10k1(1)k1)ak1++(10+1)a1+(11)a0]+(1)kak+(1)k1ak1++(1)1a1+(1)0a0 \begin{aligned}N & = [ (10^k - (-1)^k) a_k + ( 10^{k-1} - (-1)^{k-1}) a_{k-1} + \ldots + (10+1)a_1 + (1-1)a_0 ]\\& \quad + (-1)^k a_k + (-1)^{k-1}a_{k-1} + \ldots + (-1)^1 a_1 + (-1)^0 a_0\end{aligned}

Since the terms in the square brackets consist of multiples of 11, it follows that N N is a multiple of 11 if and only if the alternating sum is a multiple of 11.

Application and Extentions

Find all possible values of a a such that the number 98a6 \overline{98a6} is a multiple of 3.

From the rules of divisibility, the number 98a6 \overline{98a6} is a multiple of 3 if and only if the sum of the digits 9+8+a+6=23+a 9 + 8 + a + 6 = 23+a is a multiple of 3. Since 0a9 0 \leq a \leq 9, this implies that a=1,4,7 a = 1, 4, 7 are all the possible values.

 

Show that if the last 3 digits of a number N N are abc \overline{abc}, then N N is a multiple of 8 if and only if 4a+2b+c 4a + 2b + c is a multiple of 8.

This follows because 100a+10b+c=8(12a+b)+4a+2b+c 100a + 10 b + c = 8 (12a + b) + 4a + 2b + c. Hence, by the divisibility rule of 8, N N is a multiple of 8 if and only if abc \overline{abc} is a multiple of 8 if and only if 4a+2b+c 4a+2b+c is a multiple of 8.

 

Divisibility rule of 7: Break up the number into blocks of 6, starting from the right. Add up all these blocks, and the resultant number has to be a multiple of 7.

This follows because 1061=999999=142857×7 10^6 -1 = 999999 = 142857 \times 7, so 106kMk+106(k1)Mk1++106M1+M0 10^{6k}M_k + 10^{6(k-1)}M_{k-1} + \ldots + 10^6 M_1 + M_0 is a multiple of 7 if and only if Mk+Mk1++M0 M_k + M_{k-1} + \ldots + M_0 is a multiple of 7. If we use 0Mi999999 0 \leq M_i \leq 999999, we get the result as stated.

Show that the 6 digit number abcdef \overline{abcdef} is a multiple of 7 if and only if 5a+4b+6c+2d+3b+f 5a + 4b + 6c + 2d + 3b + f is a multiple of 7.

Solution: This follows because 1=0×7+110=1×7+3\ 100=14×7+21000=142×7+610000=1428×7+4100000=14285×7+5. \begin{aligned} &1 & = &0 \times 7 & + 1\\ &10 & = &1 \times 7 & + 3\\\ & 100 & =& 14 \times 7 & + 2\\ &1000 & = &142 \times 7 &+ 6\\ &10000 & = &1428 \times 7 &+ 4 \\ &100000& = &14285 \times 7 &+ 5.\\ \end{aligned}

Hence abcdef \overline{abcdef} is a multiple of 7 if and only if 5a+4b+6c+2s+3e+f 5a + 4b + 6c + 2s + 3e + f is a multiple of 7.

Note by Arron Kau
5 years, 6 months ago

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For eight, I just like to use the fact that a number ...abc\overline {...abc} is a multiple of eight if

(1) aa is odd, and bc\overline {bc} is a multiple of four, but NOT eight, (bc4(mod8)\overline{bc} \equiv 4 \pmod{8}

OR

(2) when aa is even, and bc\overline {bc} is a multiple of eight.

This can be proved relatively simply using mod math. I find it very useful, because you need only to remember 2-digit multiples of 8 instead of 3-digit ones!

Nicolas Bryenton - 5 years, 2 months ago

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No!not at all.You have to only look at the last 3 digits. I don't know how did you found that you have to only look at the last 2 digits.

Anuj Shikarkhane - 5 years, 2 months ago

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I'm sorry but did you even read what I wrote 0_0

My method tells you if the last three digits are a multiple of eight, but without having to memorize all three digit multiples of 8. If you read it more carefully, it should make sense, because it indeed does make sense.

Nicolas Bryenton - 5 years, 2 months ago

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@Nicolas Bryenton I don't think so.

Anuj Shikarkhane - 5 years, 2 months ago

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@Anuj Shikarkhane Here is a proof I wrote.

Nicolas Bryenton - 5 years, 2 months ago

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Divisibility by 3: given a number in base 10 can be rewritten as:

(an10n)+(an110n1)+(an210n2)+...+a1(a_n10^n)+(a_{n-1}10^{n-1})+(a_{n-2}10^{n-2})+...+a_1

We know that 101(mod3)10\equiv{1(\mod{3})} So if we want to see if the above integer is divisible by 3, we can rewrite the integer reduced by modulo 3:

(an10n)+(an110n1)+(an210n2)+...+a1((an)+(an1)+(an2)+...+a1)(mod3)(a_n10^n)+(a_{n-1}10^{n-1})+(a_{n-2}10^{n-2})+...+a_1\equiv{((a_n)+(a_{n-1})+(a_{n-2})+...+a_1)(\mod{3})}

So we will have to find the sum of the digits. QED

William Isoroku - 3 years, 8 months ago

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Also, I would like to add that if a number is divisible by 11 then the alternating sum of its digits is itself divisible by 11. E.g 92818..... 9-2+8-1+8 = 22 and 11|22. Also, for 7 you take off the last digit of a number and subtract twice that number from the original. If the result is divisible by 7 then the original number is also divisible by 7 E.g. 259... 25 - 2(9) = 7 & 7|7

Curtis Clement - 4 years, 11 months ago

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Thanks for this

Anuj Shikarkhane - 5 years, 3 months ago

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