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2.5: Divisibility Rules

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In this section, we will explore some rules of divisibility for some positive integers.  These rules can be easily extended to all the integers by dropping the sign.

Let xZ+.

Then x=dn10n+dn110n1++d2102+d1101+d0,

which implies

x=10(dn10n1+dn110n2++d210+d1)+d0.

Thus we can express x as 10a+b, where b=d0 is the ones digit of x, and a=dn10n1+dn110n2++d210+d1.

Divisibilty by 2=21

 

Let xZ+. 2x iff 2b. In other words, 2 divides an integer iff the ones digit of the integer is either 0,2,4,6, or 8.

Proof:

Since 210, x=10a+b, and by divisibility theorem I, 2x iff 2b.

Divisibility by 5:

 

5x iff 5b. In other words, 5 divides an integer iff the ones digit of the integer is either 0, or 5.

Proof:

Since 510, x=10a+b, and by divisibility theorem I, 5x iff 5b.

Divisibility by 10:

 

10x iff 10b. In other words, 10 divides an integer iff the ones digit of the integer is 0,.

Proof:

Since 1010, x=10a+b, and by divisibility theorem I, 10x iff 10b.

Divisibility by 4=22:

 

4x iff 4d1d0.

Proof:
Let x be an integer. Then

 

x=dn10n+dn110n1++d2102+d1101+d0, which implies

x=100(dn10n2+dn110n3++d2)+10d1+d0=100(dn10n2+dn110n3++d2)+d1d0.

Since 4100, x=100a+d1d0, and by divisibility theorem I, 4x iff 4d1d0.

Divisibility by 4 of a two digit number

4d1d0 iff 42d1+d0

 

Divisibility by 8=23:

 

8x iff 8d2d1d0.

Proof:

Let x be an integer. Then

x=dn10n+dn110n1++d2102+d1101+d0, which implies

x=103(dn10n3+dn110n4++d3)+100d2+10d1+d0=103(dn10n3+dn110n4++d3)+d2d1d0.

Since 81000, x=1000a+d2d1d0, and by divisibility theorem I, 8x iff 8d2d1d0.

A similar argument can be made for divisibility by 2n, for any positive integer n.

Example 2.5.1:

Using divisibility tests, check if the number 824112284 is divisible by:

  1. 5
  2. 4
  3. 8

Solution:

  1.   824112284 is not divisible by 5.

Rule: The one's digit of the number has to be either a 0 or a 5.

Since the last digit is not 0 or 5,  824112284  is not divisible by 5.

2. 824112284 is divisible by 4.

Rule: The last two digits of the number have to be divisible by 4.

The last two digits of 824112284 is 84=(4)(21).

Since 84 is divisible by 4,824112284 is divisible by 4 also.

3. 824112284 is not divisible by 8.

Rule: The last three digits of the number have to be divisible by 8.

The last three digits of 824112284 is 284. Since 284 is not divisible by 8,  the original number, 824112284 is not divisible by 8 either.

Notice the following pattern 101=9,1001=99,10001=999,,10n1=9999

Divisibility by 3=31:

 

3x iff 3 divides sum of its digits.

Proof:
Let xZ+.

 

Then x=dn10n+dn110n1++d2102+d1101+d0,

which implies

x=9(dn(10n1+dn110n2++d210+d1)+(d0.

Example 2.5.2:

Find the possible values for the missing digit x, if 1234x51234 is divisible by 3.

Solution:

Consider the following:

The divisibility rule for the number 3 is as follows: If the sum of the digits in the whole number is a number divisible by 3, then the larger, original number is also.

The sum of the digits is 2(1+2+3+4)+5+x=25+x. Since 3(25+x), 25+x=27,30 or 33. Hence x=2,5 or 8.

 

Divisibility by 9=32:

 

9x iff 9 divides sum of its digits.

Divisibility by 7:

 

7x iff 7 divides the absolute difference between a2b, where x=10a+b, where b=d0 is the ones digit of x, and a=dn10n1+dn110n2++d210+d1.

Proof:

Assume 7x. Which implies 710a+b.

Consider x21b=(10a+b)21b=10a20b=10(a2b).

Thus 7x if and only if  7a2b.

Divisibility by 11:

 

11x iff 11 divides the absolute difference between the alternate sum.

Proof:
 

This page titled 2.5: Divisibility Rules is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Pamini Thangarajah.

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