The Proof of Divisibility By 5

This is the fourth post in the divisibility series for middle school students. In this post, I am going to discuss divisibility by 5. This is probably one of the easiest divisibility rules to remember and to prove. I am sure you are all familiar that a number ends in 0 or 5, then it is divisible by 5. Therefore, we have two cases.

Theorem

If a number ends in 0 or 5, then it is divisible by 5.

Proof

Numbers Ending in 0

All numbers ending in 0 can be expressed as 10n where n is some numbers. For example, 30 can be expressed as 10(3) and 120 can be expressed as 10(12). Now, since 10 is divisible by 5, so 10n is divisible by 5. This means that all numbers ending in 0 are divisible by 5. 

Numbers Ending in 5

Numbers Ending in 5 can be expressed as 10n + 5. For example, 85 can be expressed as 10(8) + 5 and 225 can be expressed as 22(10) + 5. From above, we already proved that all numbers of the form 10n is divisible by 5, so we are only left with 5. Now, since 5 is divisible by 5, 10n + 5 is divisible by 5. In fact, we can easily see that 10n is divisible by 5 because we can factor out 5 as in

10n + 5 = 5(2n + 1)

What about the numbers ending in other digits?

The numbers ending in 1, 2, 3, and 4 can be expressed as

10n + 1

10n + 2

10n + 3

10n + 4

Now, we know that 10n is divisible by 5, so we can ignore it. This leaves with 1, 2, 3, and 4 which are not divisible by 5.

The other remaining digits are 6, 7, 8 and 9. Notice that we can isolate 10n + 5 as shown in the expressions below. Since 10n + 5 is divisible by 5, again we are left with 1, 2, 3, and 4 which are not divisible by 5.

10n + 6 = (10n + 5 )+ 1

10n + 7 = (10n + 5) + 2

10 n + 8 = (10n + 5) + 3

10 n + 9 = (10n + 5) + 4

This proves that the numbers ending in 6, 7, 8, and 9 are not divisible by 5.

There you go, we have now proved that a number is divisible by 5 if it ends in 0 and it ends in 5.

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