Wednesday, September 14, 2005
Number Divisibility Tests -- Simplifying Division
While division by small numbers is not very difficult, dividing into large numbers can take time that, for instance, could waste time on a timed test. A good example is reducing fractions to lowest terms.
For example, let's reduce 72/504.
The first thing you should notice is that both terms are even. That means each term is divisible by 2. So, divide each term by 2 to get 36/252. Again, they are even. Divide by 2 again and get 18/126. Even again! Divide by 2 to get 9/63. Now you can try to divide 63 by 9 and get 7. So the answer is 1/7.
If you're real good at long division, you may have just divided 504 by 72 to begin with. Remember, this is an example.
I might have tried to divide 504 by 12, which is 42, giving 6/42 and then 1/7. But, since I know some of these divisibility tests, I would have noticed that 7+2=9 which is divisible by 3 and 5+0+4=9 also. I would have divided by 3 first giving 24/168. 2+4=6 and 1+6+8=15 both divisible by 3 and reduced the fraction to 8/56 and then 1/7. Even faster, since 72/504 is divisible by 3 and by 2, it must be divisible by 6, which gives 12/84 right away. From your 12 times table, you know 84 = 12 X 7 and gives you 1/7 quickly. Notice that 1+2=3 and 8+4=12, both divisible by 3 giving 4/28. Again, this is an example to make you think. Whatever is easiest for you is the way to go but always be open to new ideas and think outside the box.
Using some of the following divisibility tests may simplify the process.
An integer is divisible by:
2 if and only if the last digit of the number is divisible by 2.
3 if and only if the sum of its digits is divisible by 3.
4 if and only if the number formed by its last two digits is divisible by 4.
5 if and only if the last digit of the number is 0 or 5.
6 if and only if the number is divisible by 2 and by 3.
7 if and only if a new number formed by cycling the pattern {1, 3, 2, -1, -3, -2}multiplied by the digits in reverse order is divisible by 7. WHEW! Forget this one. It's easier to just divide by 7. An example is 175 is divisible by 7 since: Make a new number 571 (the reverse) and apply the pattern so that 1X5 + 3X7 + 2X1 = 5 + 21 + 2 = 28, which is divisible by 7.
8 if and only if the number formed by its last three digits is divisible by 8. Note that 24 is divisible by 8 since 024 is divisible by 8.
9 if and only if the sum of its digits is divisible by 9.
10 if and only if the last digit of the number is 0.
11 if and only if the sum of the digits in the odd-numbered places diminished by the sum of the digits in the even-numbered places is divisible by 11. Example, 121 is divisible by 11 since 1 + 1 - 2 = 0 is divisible by 11.
12 if and only if the number is divisible by 3 and by 4.
For example, let's reduce 72/504.
The first thing you should notice is that both terms are even. That means each term is divisible by 2. So, divide each term by 2 to get 36/252. Again, they are even. Divide by 2 again and get 18/126. Even again! Divide by 2 to get 9/63. Now you can try to divide 63 by 9 and get 7. So the answer is 1/7.
If you're real good at long division, you may have just divided 504 by 72 to begin with. Remember, this is an example.
I might have tried to divide 504 by 12, which is 42, giving 6/42 and then 1/7. But, since I know some of these divisibility tests, I would have noticed that 7+2=9 which is divisible by 3 and 5+0+4=9 also. I would have divided by 3 first giving 24/168. 2+4=6 and 1+6+8=15 both divisible by 3 and reduced the fraction to 8/56 and then 1/7. Even faster, since 72/504 is divisible by 3 and by 2, it must be divisible by 6, which gives 12/84 right away. From your 12 times table, you know 84 = 12 X 7 and gives you 1/7 quickly. Notice that 1+2=3 and 8+4=12, both divisible by 3 giving 4/28. Again, this is an example to make you think. Whatever is easiest for you is the way to go but always be open to new ideas and think outside the box.
Using some of the following divisibility tests may simplify the process.
An integer is divisible by:
2 if and only if the last digit of the number is divisible by 2.
3 if and only if the sum of its digits is divisible by 3.
4 if and only if the number formed by its last two digits is divisible by 4.
5 if and only if the last digit of the number is 0 or 5.
6 if and only if the number is divisible by 2 and by 3.
7 if and only if a new number formed by cycling the pattern {1, 3, 2, -1, -3, -2}multiplied by the digits in reverse order is divisible by 7. WHEW! Forget this one. It's easier to just divide by 7. An example is 175 is divisible by 7 since: Make a new number 571 (the reverse) and apply the pattern so that 1X5 + 3X7 + 2X1 = 5 + 21 + 2 = 28, which is divisible by 7.
8 if and only if the number formed by its last three digits is divisible by 8. Note that 24 is divisible by 8 since 024 is divisible by 8.
9 if and only if the sum of its digits is divisible by 9.
10 if and only if the last digit of the number is 0.
11 if and only if the sum of the digits in the odd-numbered places diminished by the sum of the digits in the even-numbered places is divisible by 11. Example, 121 is divisible by 11 since 1 + 1 - 2 = 0 is divisible by 11.
12 if and only if the number is divisible by 3 and by 4.
