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08 Sep 2013, 11:01
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Hey guys,
I discovered a really cool divisibility by 4. The general rule I have seen out there for divisibility by 4 is that the number be divided by 2 twice. Others I have seen are also about the last two digits of a number. But what if you don't know the multiplication table up to 20 for the number 4. There has to be a dumb-a** way to the divisibility by 4.
So lets start by listing some divisible numbers of 4. Let's start with 16.
16
____
20
24
28
___
32
36
___
40
44
48
___
52
56
__
60
64
68
__
72
76
__
80
84
88
__
92
96
...
With the breaks I added do you see the pattern? There are more numbers divisible by 4 that have starting even digits than odd digits-- the pattern is 3,2,3,2,3,2. All the numbers with starting odd digits ALWAYS have a 2 and a 6 on the end! All the numbers with even digits have a 0,4,8 on the end. So all you have to remember is if the starting digit is odd is to check whether their is either a 2 or a 6 for the number to be divisible by 4. This property extend to as many digits numbers.
Ex. Is 7,546,264,748,296 divisible by 4?
Well.... (7,546,264,748,29) is an odd number and the last digit is 6, hence yes it divisible by 4.
Even simpler. Just grab the last two digits. 96. Is 9 odd? Yes. Does the number end with a 2 or 6? Yes. Hence, divisible by 4.
Hope this divisibility trick helps.
I discovered a really cool divisibility by 4. The general rule I have seen out there for divisibility by 4 is that the number be divided by 2 twice. Others I have seen are also about the last two digits of a number. But what if you don't know the multiplication table up to 20 for the number 4. There has to be a dumb-a** way to the divisibility by 4.
So lets start by listing some divisible numbers of 4. Let's start with 16.
16
____
20
24
28
___
32
36
___
40
44
48
___
52
56
__
60
64
68
__
72
76
__
80
84
88
__
92
96
...
With the breaks I added do you see the pattern? There are more numbers divisible by 4 that have starting even digits than odd digits-- the pattern is 3,2,3,2,3,2. All the numbers with starting odd digits ALWAYS have a 2 and a 6 on the end! All the numbers with even digits have a 0,4,8 on the end. So all you have to remember is if the starting digit is odd is to check whether their is either a 2 or a 6 for the number to be divisible by 4. This property extend to as many digits numbers.
Ex. Is 7,546,264,748,296 divisible by 4?
Well.... (7,546,264,748,29) is an odd number and the last digit is 6, hence yes it divisible by 4.
Even simpler. Just grab the last two digits. 96. Is 9 odd? Yes. Does the number end with a 2 or 6? Yes. Hence, divisible by 4.
Hope this divisibility trick helps.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
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08 Sep 2013, 11:07
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alphabeta1234
There is a much easier way to check divisibility by 4:
If the last two digits form a number divisible by 4, the number is also. For, example 123, 456, 789,036 IS divisible by 4, since the last two digits 36 IS divisible by 4.
Similar logic applies to 8, 16, ...
For example, if the last three digits of a number are divisible by 8, then so is the whole number.
For more check here: math-number-theory-88376.html
Hope it helps.
finished
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10 Sep 2013, 19:49
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Your making this way too difficult.. its a simple rule.. if you can divide the last two numbers twice and its a whole number, its divisible by 4
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
