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 x∈Z+.
Then x=dn10n+dn−110n−1+⋯+d2102+d1101+d0,
which implies
x=10(dn10n−1+dn−110n−2+⋯+d210+d1)+d0.
Thus we can express x as 10a+b, where b=d0 is the ones digit of x, and a=dn10n−1+dn−110n−2+⋯+d210+d1.
Divisibilty by 2=21:
Let x∈Z+. 2∣x iff 2∣b. In other words, 2 divides an integer iff the ones digit of the integer is either 0,2,4,6, or 8.
- Proof:
-
Since 2∣10, x=10a+b, and by divisibility theorem I, 2∣x iff 2∣b.◻
Divisibility by 5:
5∣x iff 5∣b. In other words, 5 divides an integer iff the ones digit of the integer is either 0, or 5.
- Proof:
-
Since 5∣10, x=10a+b, and by divisibility theorem I, 5∣x iff 5∣b.◻
Divisibility by 10:
10∣x iff 10∣b. In other words, 10 divides an integer iff the ones digit of the integer is 0,.
- Proof:
-
Since 10∣10, x=10a+b, and by divisibility theorem I, 10∣x iff 10∣b.◻
Divisibility by 4=22:
4∣x iff 4∣d1d0.
- Proof:
- Let x be an integer. Then
x=dn10n+dn−110n−1+⋯+d2102+d1101+d0, which implies
x=100(dn10n−2+dn−110n−3+⋯+d2)+10d1+d0=100(dn10n−2+dn−110n−3+⋯+d2)+d1d0.
Since 4∣100, x=100a+d1d0, and by divisibility theorem I, 4∣x iff 4∣d1d0.◻
Divisibility by 4 of a two digit number
4∣d1d0 iff 4∣2d1+d0
Divisibility by 8=23:
8∣x iff 8∣d2d1d0.
- Proof:
-
Let x be an integer. Then
x=dn10n+dn−110n−1+⋯+d2102+d1101+d0, which implies
x=103(dn10n−3+dn−110n−4+⋯+d3)+100d2+10d1+d0=103(dn10n−3+dn−110n−4+⋯+d3)+d2d1d0.
Since 8∣1000, x=1000a+d2d1d0, and by divisibility theorem I, 8∣x iff 8∣d2d1d0.◻
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:
- 5
- 4
- 8
Solution:
- 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 10−1=9,100−1=99,1000−1=999,⋯,10n−1=999⋯9
Divisibility by 3=31:
3∣x iff 3 divides sum of its digits.
- Proof:
- Let x∈Z+.
Then x=dn10n+dn−110n−1+⋯+d2102+d1101+d0,
which implies
x=9(dn(10n−1+dn−110n−2+⋯+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:
9∣x iff 9 divides sum of its digits.
Divisibility by 7:
7∣x iff 7 divides the absolute difference between a−2b, where x=10a+b, where b=d0 is the ones digit of x, and a=dn10n−1+dn−110n−2+⋯+d210+d1.
- Proof:
-
Assume 7∣x. Which implies 7∣10a+b.
Consider x−21b=(10a+b)−21b=10a−20b=10(a−2b).
Thus 7∣x if and only if 7∣a−2b.
Divisibility by 11:
11∣x iff 11 divides the absolute difference between the alternate sum.
- Proof: