The ability to find factors of a number helps simplify division problems
and increase the speed an accuracy of solving multiplication problems.
Sometimes it's imperative to be able to factor a two or three digit
number or higher in the middle of a multiplication problem. The rules in
the upcoming paragraphs will help in this process. You could also
quickly tell someone that 57,231 is divisible by 3 and 9. Simple rules
will make people think you're a math genius, when in fact, any average
person can learn these divisibility rules.
Let's start with the easier rules. To check if a number is divisible
by 2 is to simply determine whether the number is even or odd. Any
number ending in 0, 2, 4, 6, or 8, therefore any even number is
divisible by two, any odd number isn't.
Also very simple is divisibility by 5. Any number ending in 0 or 5 is
divisible by 5. This is obvious since the multiples of 5 (0, 5, 10, 15,
20, 25, 30, ....) all end in either 0 or 5.
Divisibility by 10 may be the simplest of all. Any number ending in 0 is divisible by 10. No other number is divisible by 10.
To see if a number is divisible by 4, check the last two digits of the
number. If this two digit number is divisible by 4, the entire number
is. For example, the number 16,232 is divisible by 4 since 32 is
divisible by 4. If a number ends in 00, it is also divisible by 4 since
100, 200, 300, 400, 500, 600, 700, 800, and 900 are all divisible by 4.
Checking if a number is divisible by 8 is similar to the previous
method for divisibility by 4. Except now we check the last 3 digits of
the number. If the 3 digit number is divisible by 8, the entire number
is. For example, 66,512 is divisible by 8 because 512 divided by 8 is
64. This works because 1,000 and any thousand is divisible by 8.
The test for divisibility by 3 is also quite simple. Add the digits of
a number. If the sum is divisible by 3, the entire number is divisible
by 3. In the example in the first paragraph, add the digits of 57,231.
The sum of the numbers is 5 + 7 + 2 + 3 + 1 = 18. We know that 18 is
divisible by 3, so 57,231 is divisible by 3. Likewise, the same kind of
rule holds true for divisibility by 9. If the sum of the digits is
divisible by 9, the entire number is divisible by 9. We know 18 is
divisible by 9, so is 57,231.
Testing for divisibility of 6 is simple once we know how to determine
if a number is divisible by 3. If a number is even and divisible by 3,
then it's also divisible by 6. Therefore 3,168 is divisible by 6, but
2,115 is not.
Another interesting test is for divisibility by 11. Take the number
you are testing and alternately subtract and add the digits. If the
result is 0 or a multiple of 11, then the number is divisible by 11. For
example, take 36,713 and apply the rule (3 - 6 + 7 -1 + 3 = 6).
Therefore 36,713 is not divisible by 11. Now apply the rule to 619,091.
We get 6 - 1 + 9 - 0 + 9 - 1 = 22. Since 22 is divisible by 11, so is
619,091.
Now we get to the most difficult divisibility rule. To test if a
number is divisible by 7, add or subtract a number that is a multiple of
7 to the number you are checking. Try to add or subtract a multiple of 7
so the number you get ends in 0. For example, suppose you want to know
if 6,358 is divisible by 7. If we add 42, which is a multiple of 7, we
get 6,400. We can remove the 0's since we know dividing by 10 doesn't
affect the divisibility of 7. Now we have 64. We know 64 is not
divisible by 7, therefore 6,358 is not divisible by 7.
Let's try one more number to test for divisibility by 7. Suppose we
have 1,106. If we add 14, we get 1,120. Drop the 0 to get 112. Now we
can add 28 (another multiple of 7) to get 140. We know 140 is divisible
by 7 and if not, we drop the 0 to get 14, which is clearly divisible by
7. Therefore 1,106 is divisible by 7.
I have taught many students these rules over the past 14 years, some
come in handy, others are just very interesting bits of information to
the math lover. For all practical purposes we can check if a number is
divisible by another number by simply using our handy calculators. But
for those interested in math, want to get better at math or simply want
to impress someone with quick mental math, these divisibility rules will
come in handy.
Thursday, March 27, 2014
Monday, March 24, 2014
Early in elementary school students are taught how to subtract single
digit numbers. Then students progress to subtraction of two digit
numbers, which is taught from right to left. For many people, it's
easier to add than subtract, but if you use the same left to right
method of subtraction like I showed in addition in a previous article,
subtraction can be almost as simple as addition.
To subtract two digit numbers mentally from left to right, you want to try to simplify the problem so that you are subtracting or adding a one digit number. Let's start with a simple example, 75 minus 23. The idea is to make each subtraction easier until you get the final answer. Break down the second number to a number in base 10 and another number. So break down 23 to 20 and 3. Now take 75 minus 20 to get 55 and subtract 3 to get the final answer of 52. A problem without borrowing is very simple, whether subtracting from left to right or right to left.
Let's try another example, this time with borrowing. For example, 55 minus 37. In this problem we have to borrow. This occurs when the larger number is being subtracted from the smaller number. There are two different ways to approach such problems. First, you can change 37 to 30 and 7 and subtract 30 from 55 first to get 25 then subtract 7 to get 18. But in this case, it's easier to subtract in a different way. Change 37 to 40 minus 3. By doing this we subtract 40 from 55 to get 15 and add back the 3 to get 18.
The easy way to decide which method to use is to see if you have to borrow or not. If there is borrowing involved, round the number you are subtracting up to the next 10, subtract that and add back the difference. Notice that's what I did in the above example. I subtracted 40 from 55 to get 15 and added back the 3 to get 18. If there is no borrowing involved, round the number down and subtract all the way through.
Let's try another example. Take 47 minus 29. Since there is borrowing involved, change 29 to 30 minus 1. Now subtract 30 from 47 to get 17 and add 1 back to get 18. With a little practice, you'll be able to solve subtraction problems using both methods.
When subtracting a three digit number from another three digit number, you used the same principles. If there is no borrowing involved the subtraction is easily done from left to right. Take 645 minus 124. Change 124 to 100 plus 20 plus 4. Now subtract 100 from 645 to get 545. Subtract 20 from 545 to get 525 and finally subtract 4 to get 521.
Take an example of subtracting three digit numbers with borrowing. Try 423 minus 219. We can either change 219 to 200 plus 10 plus 9 and subtract to get 423 minus 200 equals 223, 223 minus 10 equals 213, and 213 minus 9 equals 204. Or we can change 219 to 300 minus 81. Now we subtract 300 from 423 to get 123 and add back 81 to get 204. The choice on which method you use is entirely up to you. Both will work nicely, it's a matter or personal preference.
These are some techniques I have used over the years as a math tutor to help encourage students to perform more math mentally instead of relying on a calculator. There will be more mental math articles to come. Enjoy!
To subtract two digit numbers mentally from left to right, you want to try to simplify the problem so that you are subtracting or adding a one digit number. Let's start with a simple example, 75 minus 23. The idea is to make each subtraction easier until you get the final answer. Break down the second number to a number in base 10 and another number. So break down 23 to 20 and 3. Now take 75 minus 20 to get 55 and subtract 3 to get the final answer of 52. A problem without borrowing is very simple, whether subtracting from left to right or right to left.
Let's try another example, this time with borrowing. For example, 55 minus 37. In this problem we have to borrow. This occurs when the larger number is being subtracted from the smaller number. There are two different ways to approach such problems. First, you can change 37 to 30 and 7 and subtract 30 from 55 first to get 25 then subtract 7 to get 18. But in this case, it's easier to subtract in a different way. Change 37 to 40 minus 3. By doing this we subtract 40 from 55 to get 15 and add back the 3 to get 18.
The easy way to decide which method to use is to see if you have to borrow or not. If there is borrowing involved, round the number you are subtracting up to the next 10, subtract that and add back the difference. Notice that's what I did in the above example. I subtracted 40 from 55 to get 15 and added back the 3 to get 18. If there is no borrowing involved, round the number down and subtract all the way through.
Let's try another example. Take 47 minus 29. Since there is borrowing involved, change 29 to 30 minus 1. Now subtract 30 from 47 to get 17 and add 1 back to get 18. With a little practice, you'll be able to solve subtraction problems using both methods.
When subtracting a three digit number from another three digit number, you used the same principles. If there is no borrowing involved the subtraction is easily done from left to right. Take 645 minus 124. Change 124 to 100 plus 20 plus 4. Now subtract 100 from 645 to get 545. Subtract 20 from 545 to get 525 and finally subtract 4 to get 521.
Take an example of subtracting three digit numbers with borrowing. Try 423 minus 219. We can either change 219 to 200 plus 10 plus 9 and subtract to get 423 minus 200 equals 223, 223 minus 10 equals 213, and 213 minus 9 equals 204. Or we can change 219 to 300 minus 81. Now we subtract 300 from 423 to get 123 and add back 81 to get 204. The choice on which method you use is entirely up to you. Both will work nicely, it's a matter or personal preference.
These are some techniques I have used over the years as a math tutor to help encourage students to perform more math mentally instead of relying on a calculator. There will be more mental math articles to come. Enjoy!
Thursday, March 20, 2014
Tuesday, March 18, 2014
Here's a fun math secret one can do with their friends to convince them
that you are a human calculator. When in fact, the method behind the
calculation is quite simple. First, ask a friend to pick a two digit
number and cube it. For example, suppose the volunteer selects 52. He
cubes the number to get 52 times 52 times 52 equals 140,608. Then ask
the volunteer to give the number. Very quickly and almost instantly with
practice, you can say that the cube root is 52. How is this done?
First you must know the cubes from 1 to 10. The first 10 cubes are 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000. Now that you know the first 10 cubes, here is the method.
Look at the magnitude of the thousands digits. Since the number is 140,608, the thousands digits are 140. Now think of what the cube root of 140 is. We know it's between 5 and 6 since 5 cubed is 125 and 6 cubed is 216. So the first number in the answer is 5. Now look at the last number, which is 8. All we do is see which cube ends in 8. See that 2 cubed is 8 so the answer is 52. Notice the cubes from 1 to 10 end in the numbers 0 through 9, no number is repeated twice. There lies the key to solving these cube roots.
Let's try another example. What is the cube root of 54,872? The thousands digits in this problem are 54. We know the cube root of 54 falls between 3 and 4 since 3 cubed is 27 and 4 cubed is 64. So we know the answer starts with 3. Next look at the last number, which is 2. Which of the first ten cubes end in 2? Looking at the list, 8 cubed is 512, so 8 is the last number. Therefore, the answer is 38.
Note that this method will not work unless the cube root is a whole number.
Over my years of tutoring math, I have used these tricks that I have learned with my students.
Practice calculating cube roots in this fashion and you can wow your friends. Even if you are not particularly a math wizard, you will appear to be one.
First you must know the cubes from 1 to 10. The first 10 cubes are 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000. Now that you know the first 10 cubes, here is the method.
Look at the magnitude of the thousands digits. Since the number is 140,608, the thousands digits are 140. Now think of what the cube root of 140 is. We know it's between 5 and 6 since 5 cubed is 125 and 6 cubed is 216. So the first number in the answer is 5. Now look at the last number, which is 8. All we do is see which cube ends in 8. See that 2 cubed is 8 so the answer is 52. Notice the cubes from 1 to 10 end in the numbers 0 through 9, no number is repeated twice. There lies the key to solving these cube roots.
Let's try another example. What is the cube root of 54,872? The thousands digits in this problem are 54. We know the cube root of 54 falls between 3 and 4 since 3 cubed is 27 and 4 cubed is 64. So we know the answer starts with 3. Next look at the last number, which is 2. Which of the first ten cubes end in 2? Looking at the list, 8 cubed is 512, so 8 is the last number. Therefore, the answer is 38.
Note that this method will not work unless the cube root is a whole number.
Over my years of tutoring math, I have used these tricks that I have learned with my students.
Practice calculating cube roots in this fashion and you can wow your friends. Even if you are not particularly a math wizard, you will appear to be one.
Friday, March 14, 2014
Remember when graphing the curves of trig functions sine and cosine, need to know the graph of the basic functions. Then look for amplitude, phase shift, vertical shift and spacing.
Sinx has period 2Pi, as does Cosx
aSin(bx + c) + d
a is the amplitude
2Pi/b is the period
c is the phase shift
d is the vertical shift
To get the spacing, divide the period by 4
Remember the parent functions sine and cosine have max of 1 and min of -1.
Sinx has period 2Pi, as does Cosx
aSin(bx + c) + d
a is the amplitude
2Pi/b is the period
c is the phase shift
d is the vertical shift
To get the spacing, divide the period by 4
Remember the parent functions sine and cosine have max of 1 and min of -1.
Monday, March 10, 2014
Some mental math
For those that love math and how numbers are manipulated using mathematical
operations, this article is for you. For those that want to improve
their mental math skills, this article is for you as well. Even those
that don't like math can benefit from learning how to do mental math.
There are many types of problems one can do mentally, but for the scope
of this article, I will concentrate on the simple problem of multiplying
any two-digit number by 11.
As a math tutor for 14 years, I have taught several students this method in attempts to transition from the use of a calculator for all problems to performing more calculations in the head.
Let's take a simple example, multiplying 34 by 11. To solve the problem, simply add the digits 3 + 4 = 7 and write the digits between the 3 and the 4. Your answer is 374. Very simply done. You can check your answer by multiplying by hand or using a calculator. Let's take another example, 42 times 11. Add the 4 +2 = 6 and write 6 between the 4 and 2. The answer is 462. Now that was easy, but don't get overly excited just yet. There are other cases where a little more needs to be done.
For example, suppose the problem is 67 times 11. Although 6 + 7 = 13, the answer is not 6,137. As before the 3 goes in between the numbers, but the 1 needs to be added to the 6 to get 7. The correct answer is 737. Basically think of adding the 1 to the 6 as "carrying" the 1 like you would in a simple problem such as 13 + 8, add the 3 and 8 to get 11, write down the 1 and carry the 1, adding it to the other 1 to get 21. Let's try another example, suppose the problem is 84 times 11. Add 8 and 4 to get 12. Write the 2 in between the 8 and 4 but remember to carry the 1, adding to the 8 to get 924. See how easy that is?
This is just one of several articles I plan to write on mental math. Others will include mental addition and subtraction, basic multiplication, mental division, estimations and advanced multiplication. In the meantime, practice the mental math you learned in this article. It will exercise your mind and impress your friends!
As a math tutor for 14 years, I have taught several students this method in attempts to transition from the use of a calculator for all problems to performing more calculations in the head.
Let's take a simple example, multiplying 34 by 11. To solve the problem, simply add the digits 3 + 4 = 7 and write the digits between the 3 and the 4. Your answer is 374. Very simply done. You can check your answer by multiplying by hand or using a calculator. Let's take another example, 42 times 11. Add the 4 +2 = 6 and write 6 between the 4 and 2. The answer is 462. Now that was easy, but don't get overly excited just yet. There are other cases where a little more needs to be done.
For example, suppose the problem is 67 times 11. Although 6 + 7 = 13, the answer is not 6,137. As before the 3 goes in between the numbers, but the 1 needs to be added to the 6 to get 7. The correct answer is 737. Basically think of adding the 1 to the 6 as "carrying" the 1 like you would in a simple problem such as 13 + 8, add the 3 and 8 to get 11, write down the 1 and carry the 1, adding it to the other 1 to get 21. Let's try another example, suppose the problem is 84 times 11. Add 8 and 4 to get 12. Write the 2 in between the 8 and 4 but remember to carry the 1, adding to the 8 to get 924. See how easy that is?
This is just one of several articles I plan to write on mental math. Others will include mental addition and subtraction, basic multiplication, mental division, estimations and advanced multiplication. In the meantime, practice the mental math you learned in this article. It will exercise your mind and impress your friends!
Sunday, March 9, 2014
Recently posted this on another site but thought worth posting here.
When solving to for the sides of a right triangle, we use the Pythagorean Theorem, where a2 +b2 = c2. Once we know the sides, we can use trigonometric functions sine, cosine and tangent to find the measures of the other two angles. An oblique triangle is one that is not right. But how does one solve for the sides and angles of an oblique triangle? The Law of Sines is used if a side and two angles are known, two angles and the sides between them are known, or two sides and an angle opposite one side is known. The formula is very easy, sinA/a = sinB/b = sinC/c. In the case of two sides and included angle or all three sides given, the Law of Cosines must be used. What is the Law of Cosines?
The Law of Cosines may look confusing at first, but is quite simple to use and memorize. If A, B, and C are the measures of the angles of a triangle, and a, b and c are the lengths of the sides opposite of those angles, then a2 = b2 + c2- 2bccosA, b2 = a2 + c2 - 2accosB and c2 = a2 + b2 - 2abcosC. Notice the side we are solving for and the angle we take the cosine of. It's the angle opposite the side. The other two sides are the ones we square on the other side of the equation and multiplied together by 2.
To solve a SAS triangle, first use the Law of Cosines to find the side opposite the angle given. Then use the Law of Sines to find the angle opposite the shorter of the two given sides. Then find the third angle by subtracting the sum of the measures of the two angles from 180.
Example: Solve the triangle with A = 50 degrees, b = 15, c = 20.
Using Law of Cosines we get a2 = (15)2 + (20)2 - 2(15)(20)cos(50)
= 225 + 400 - 600cos(50)
= 625 - 385.67
= 239.33
a = 15.47
Find angle B using Law of Sines.
SinB/15 = Sin(50)/15.47
SinB = 0.7
B = 48 degrees
Angle C is 180 - (50 + 48) = 180 - 98 = 82.
To solve a SSS triangle, use the Law of Cosines to find the angle opposite the longest side. Then use the Law of Sines to find either of the other remaining angles. Then subtract the sum of the other two angles from 180.
Example: Solve the triangle if a = 6, b = 10, c = 12.
Use the Law of Cosines to find angle C.
c2 = a2 + b2 - 2abcosC
(12)2= (6)2 + (10)2 - 2(6)(10)cosC
144 = 36 + 100 - 120cosC
144 = 136 - 120cosC
8 = -120cosC
-0.066 = cosC
C = 93.8
Use Law of Sines to get angle A
SinA/6 = Sin(93.8)/12
SinA/6 = 0.083
SinA = 0.499
A = 29.9 degrees
B = 180 - (93.8 + 29.9) = 180 - 123.7 = 56.3 degrees
Notice how the Law of Cosines always solves for a part of the triangle which enables you to use the Law of Sines. I've been using example such as the one's above during my 14 years of tutoring trigonometry. They should help any student who is having difficulty using the Law of Cosines.
When solving to for the sides of a right triangle, we use the Pythagorean Theorem, where a2 +b2 = c2. Once we know the sides, we can use trigonometric functions sine, cosine and tangent to find the measures of the other two angles. An oblique triangle is one that is not right. But how does one solve for the sides and angles of an oblique triangle? The Law of Sines is used if a side and two angles are known, two angles and the sides between them are known, or two sides and an angle opposite one side is known. The formula is very easy, sinA/a = sinB/b = sinC/c. In the case of two sides and included angle or all three sides given, the Law of Cosines must be used. What is the Law of Cosines?
The Law of Cosines may look confusing at first, but is quite simple to use and memorize. If A, B, and C are the measures of the angles of a triangle, and a, b and c are the lengths of the sides opposite of those angles, then a2 = b2 + c2- 2bccosA, b2 = a2 + c2 - 2accosB and c2 = a2 + b2 - 2abcosC. Notice the side we are solving for and the angle we take the cosine of. It's the angle opposite the side. The other two sides are the ones we square on the other side of the equation and multiplied together by 2.
To solve a SAS triangle, first use the Law of Cosines to find the side opposite the angle given. Then use the Law of Sines to find the angle opposite the shorter of the two given sides. Then find the third angle by subtracting the sum of the measures of the two angles from 180.
Example: Solve the triangle with A = 50 degrees, b = 15, c = 20.
Using Law of Cosines we get a2 = (15)2 + (20)2 - 2(15)(20)cos(50)
= 225 + 400 - 600cos(50)
= 625 - 385.67
= 239.33
a = 15.47
Find angle B using Law of Sines.
SinB/15 = Sin(50)/15.47
SinB = 0.7
B = 48 degrees
Angle C is 180 - (50 + 48) = 180 - 98 = 82.
To solve a SSS triangle, use the Law of Cosines to find the angle opposite the longest side. Then use the Law of Sines to find either of the other remaining angles. Then subtract the sum of the other two angles from 180.
Example: Solve the triangle if a = 6, b = 10, c = 12.
Use the Law of Cosines to find angle C.
c2 = a2 + b2 - 2abcosC
(12)2= (6)2 + (10)2 - 2(6)(10)cosC
144 = 36 + 100 - 120cosC
144 = 136 - 120cosC
8 = -120cosC
-0.066 = cosC
C = 93.8
Use Law of Sines to get angle A
SinA/6 = Sin(93.8)/12
SinA/6 = 0.083
SinA = 0.499
A = 29.9 degrees
B = 180 - (93.8 + 29.9) = 180 - 123.7 = 56.3 degrees
Notice how the Law of Cosines always solves for a part of the triangle which enables you to use the Law of Sines. I've been using example such as the one's above during my 14 years of tutoring trigonometry. They should help any student who is having difficulty using the Law of Cosines.
Thursday, March 6, 2014
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