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.
Monday, September 05, 2005
Methods of Proof
DIRECT PROOF:
At any stage of a proof, an implication may be replaced by its contrapositive {e.g., p = ~(~p)}. At any stage of a proof, any true expression may be replaced by another true expression {e.g., if the term sin²x is in an expression, (1 - cos²x) can be substituted}.
INDIRECT PROOF:
If ~p is false, then p is true. Therefore, to prove p is true, assume ~p is true and show that a direct proof with this assumption leads to a contradiction.
PROOF BY COINCIDENCE:
Assume two points or lines are different. Show that a direct proof with this assumption leads to a contradiction.
MATHEMATICAL INDUCTION:
If a set A contains all integers n≥a and for all n in A a proposition pn is to be proved, this can be done as follows:
(1) prove pa is true (the initial case);
(2) assume then that pk is true and prove pk--->pk+1 is true.
REVERSING STEPS:
Assume what is to be proved is already true. Perform valid operations until a result is obtained which is known to be true. Then reverse the steps (if this is possible) and the proof follows!
At any stage of a proof, an implication may be replaced by its contrapositive {e.g., p = ~(~p)}. At any stage of a proof, any true expression may be replaced by another true expression {e.g., if the term sin²x is in an expression, (1 - cos²x) can be substituted}.
INDIRECT PROOF:
If ~p is false, then p is true. Therefore, to prove p is true, assume ~p is true and show that a direct proof with this assumption leads to a contradiction.
PROOF BY COINCIDENCE:
Assume two points or lines are different. Show that a direct proof with this assumption leads to a contradiction.
MATHEMATICAL INDUCTION:
If a set A contains all integers n≥a and for all n in A a proposition pn is to be proved, this can be done as follows:
(1) prove pa is true (the initial case);
(2) assume then that pk is true and prove pk--->pk+1 is true.
REVERSING STEPS:
Assume what is to be proved is already true. Perform valid operations until a result is obtained which is known to be true. Then reverse the steps (if this is possible) and the proof follows!
Monday, August 29, 2005
Tricky Questions
Are you ready to answer some tricky, but common sense type questions? Some of these were on apptitude tests for government civil service exams. Some have been around forever. Some I made up. Have fun!
Questions 1 thru 10
Questions 11 thru 20
Questions 21 thru 30
Questions 31 thru 40
Questions 41 thru 50
Questions 51 thru 60
Questions 1 thru 10
Questions 11 thru 20
Questions 21 thru 30
Questions 31 thru 40
Questions 41 thru 50
Questions 51 thru 60
