Completing the Square
For Leading Coefficients that aren't 1

We looked at completing the square for equations who's leading coefficients are 1.  If the leading factor is not 1 we factor the number out of the leading coefficient and the second term so that the leading coefficient is one and then we proceed to complete the square.

The technique is very easy.

Lets look at a quadratic equation in the standard form.

Y=AX²+BX+C 
We factor A out of both the A and B term.
Y=A(X²+[B/A]X)+C 
Notice how we don't factor the A term out of C

To write this equation in vertex form, we take 1/2 of the B/A value and write our equation like this:

Y=(X+[1/2][A/B])²+k

It looks kind of messy but lets use an example.
Y=2X²+4X-3

Our equation now looks like this:

Y=2(X²+[4/2]X)-3 =2(X²+2X)-3
Y=2(X+1)²-3-(2)(1)²=2(X+1)²-5
Vertex is at X=-1 Y=-5 or (-1,-5)

So watch what we did with the k term.  
Lets expand 2(X+1)²

2X²+4X+2So we gained a 2 in the expansion of our modified equation so we have to also subtract a 2.
So our form becomes A(X-h)²+k
k is calculated by taking C-A(h)²

So lets do another example:
Y=-3X²-12X+4
Factor the -3 out of the first and second term:
-3(X²+4X)+4
Complete the square:
-3(X+2)²+4-(-3)(2)²=-3(X+2)²+16
The vertex is at:  X=-2  Y=16 or (-2,16)

Practice:
Y=6X²+18X+2
Y=-2X²+5X+3
Y=-7X²-21X+14

Answers posted in next post:

Completing the Square

Completing the square is a technique used to take a quadratic equation and put it in a form called the vertex form.  This is an important technique that does not stop just in algebra.  We will revisit this in the calculus post for integration techniques.

The technique is very easy.

Lets look at a quadratic equation in the standard form.

Y=X²+BX+C  

To write this equation in vertex form, we take 1/2 of the B value and write our equation like this:

Y=(X+1/2B)²+k

Both of these equations are equivalent so
We expand (X+1/2B)², we have (X+1/2B)²=X²+BX+1/4B²

We know that (X+1/2B)²-k will expand to: X²+BX+1/4B²+k=X²+BX+C 

So C=1/4B² + k or if we want to find k we do the following:
k=C-1/4B²

Not too bad right?  We are going to now do an example.

Y=X²+6X+8
B=6
1/2B=3

C=8

(X+1/2B)²=(X+3)²

k=8-3²=-1

So our equation becomes:

Y=(X+3)²-1
Vertex form is (X-h)²+k
The vertex is at x=-3 and y=-1 or (-3,-1)

Lets look at:

Y=X²-4X-3

B=-4

1/2B=-2

C=-3

(X+1/2B)²=(X-4)²
k=-3-(-2)²=-3-4=-7

So our Vertex form becomes:

Y=(X-2)²-7
Our vertex is at: (-2,-7)

Note that:   Even though our 1/2B term was negative we still ended up subtracting it from the C term.

Homework:

Factor the following into the vertex form by completing the square:

Y=X²+12X-7
Y=X²-6X+6

Y=X²+11X-1
Y=X²-14X-31
Y=X²-9x+4

The solutions are posted here.

In future posts, we will be looking at completing the square on equations with multipliers on the leading coefficient, a synthetic way of completing the square, and an explanation of why completing the square works. 

Solution Set for Basic Completion of the Square

Y=X²+12X-7

B=12
1/2B=6
(X+1/2B)²=(X+6)²

k=-7-6²=-43

Vertex form

Y=(X+6)²-43

Vertex is at: X=-6 and Y=-43

Y=X²-6X+6

B=-6
1/2B=-3
(X+1/2B)²=(X-3)²

k=6-(-3)²=-3

Vertex form

Y=(X-3)²-3

Vertex is at: X=3 and Y=-3

Y=X²+11X-1

B=11
1/2B=11.5
(X+1/2B)²=(X+11.5)²

k=-1-(11.5)²=-1-132.25=-133.25

Vertex form

Y=(X+11.5)²-133.25

Vertex is at: X=-11.5 and Y=-133.25

Y=X²-14X-31

B=-14
1/2B=-7
(X+1/2B)²=(X-7)²

k=-31-(-7)²=-31-49=-80

Vertex form

Y=(X-7)²-80

Vertex is at: X=7 and Y=-80

Y=X²-9x+4

B=-9
1/2B=-4.5
(X+1/2B)²=(X-4.5)²

k=4-(-4.5)²=4-20.25=-16.25

Vertex form

Y=(X-4.5)²-16.25

Vertex is at: X=4.5 and Y=-16.25

Simple Factoring of Quadratics

How do you factor quadratics?

We are going to learn a simple method for factoring quadratics when the leading multiplier is 1 and the quadratic has non-complex integer solutions that are positive.
If you are at this point you have learned how to FOIL if not visit the link.

Lets expand (x+8)(x+21)
(x+8)(x+21)=x²+29x+168

Now how would we factor something like this if we did not know what the factors are.
If we expand a quadratic, it looks like this:  (x+A)(x+B)=x²+(A+B)x+AB
The second term or the number before x is the addition of 2 roots or A+B
The last term is the multiplication of 2 roots or AB.

First If we factor 168 into all of its roots, we will find all potential factors:

168 factors into these pairs:     1         168
                                                 2         84
                                                 3         56
                                                 4         42
                                                 6         28
                                                 8         21
                                                 12       14
                                                 24         7

Now we see which 2 pairs add up to 29.  The only 2 that add up to 29 are 8 and 21.

Lets do a couple of simpler problems where we do not already know the factors:

Example 1:      x²+12x+32

Lets factor 32:                                                     32
                                                                          2    16
                                                                                2    8
                                                                                     2   4
                                                                                         2   2
                                                      or
                                                    1         32
                                                    2         16
                                                    4          8

So
1 and 32 add up to 33
2 and 16 add up to 18
4 and 8 add up to 12 so this is our solution set so it looks like this factored:
x²+12x+32=(x+4)(x+8)

One more example:
x²+10x+21

 Lets factor 21:                                                 1        21
                                                                         3         7

We see the only 2 number that add up to 10 are 3 and 7 so:
x²+10x+21=(x+3)(x+7)

In a future post we will do numbers with negative roots.

Pascals Triangle

Here is a Pascals triangle for up to a 20th order polynomial.  The second number in the series is the order of the polynomial.

                                                                             1

                                                                                                                    1    1

                                                                                                                  1   2   1

                                                                                                                1   3   3   1

                                                                                                            1  4    6    4   1

                                                                                                       1   5    10  10   5   1

                                                                                                     1   6   15   20   15   6   1

                                                                                                  1   7   21   35   35   21   7   1

                                                                                               1   8   28   56   70   56   28   8   1

                                                                                        1   9   36   84   126   126   84   36   9   1     

                                                                               1   10   45   120   210   252   210   120   45   10   1

                                                                          1   11   55   165   330   462   462   330   165   55   11   1  

                                                                      1   12   66   220   495   792   924   792   495   220   66   12   1

                                                                  1   13   78   286   715   1287   1716   1716   1287   715   286   78   13   1

                                                        1   14   91   364   1001   2002   3003   3432   3003   2002   1001   364   91   14   1

                                                 1   15   105   455   1365   3003   5005   6435   6435   5005   3003   1365   455   105   15   1

                                      1   16   120   560   1820   4368   8008   11440   12870   11440   8008   4368   1820   560   120   16   1

                               1   17   136   680   2380   6188   12376   19448   24310   24310   19448   12376   6188   2380   680   136   17   1

                         1   18   153   816   3060   8568 18564   31824   43758   48620   43758   31824   18564   8568   3060   816   153   18   1  
                1   19   171   969   3876   11628   27132   50388   75582   92378   92378   75582   50388   27132   11628   3876   969   171   19   1
1   20   190   1140   4845   15504   38760   77520   125970   167960   184756   167960   125970   77520   38760   15504   4845   1140   190   20   1  

The Derivative of x ^x

^{When taking the derivative of} x ^{x it often looks very hard but this is one of the functions you can always understand if you can understand how it is derived.}

^y=x x

^{The trick here is to take the ln of both sides which allows us to use the logarithmic power rule.}

lny=lnx ^x

^{Power rule of logs:}

lny=xlnx

Take the derivative of both sides and on the right side we will use the quotient rule and on the left side we will use implicit differentiation and the chain rule.

y '(1/y)=lnx +x(1/x)=lnx+1

y '=y(lnx+1)

We know y=x ^x  

                          ^so

^{y '=}x ^x(lnx+1)

Pascal's Triangle

1

1 1      1

2 1   2    1

3 1   3    3    1

4 1    4   6    4    1

5 1   5   10   10   5   1

6 1   6  15   20   15   6    1

7  1  7  21  35  35   21     7   1

On the last post we looked at binomial expansion.  It involved calculating combinations as well as carrying large numbers and was not very easy.  With Pascal's triangle we, there is no need to calculate combinations because they are given on the table.

So lets look at the table.  The numbers in red are the order of polynomial ad the number go in front of the variable from left to right.  To show this lets say that we wanted to see what (x+1)⁴:

This equation is 4th order.  So we use: 1       4       6        4         1

This becomes:   (x+1)⁴=1 x⁴   +  4 x³   +   6 x²   +    4x    +     1= x⁴+4x³+x²+4x+1

That's pretty nifty isn't it?

If we wanted (x+y)⁴=x⁴+4x³y+x²y²+4xy³+y⁴

 So you might ask if you need to memorize this table of does this table only go down to 7th order equations.  The answer is no to both questions.  You can construct this table and it goes down to as large a order as you want.

How do we construct it.

Start with                                                                                                  1

Put 1 on outside of the 1                                                                        1    1

Put one on the outside of the 1 and add the middle numbers:             1    2    1      (2=1+1)

Repeat                                                                                                1    3    3   1    (3=1+1)
Repeat                                                                                             1    4     6   4   1   (4=3+1) (6=3+3)

So if you are observing, you will see the number in the triangle is between two numbers and it is the sum of those 2 numbers.

The order of the equation starts at 1  at the two 1's ie                               1   1    1

The numbers descend from 1 from there:                                                2   1   2    1

                                                                                                              3    1    3   3     1

                                                                                                         ...................................

So just to show how easy it is if we had (x+1)⁸=x⁸+8x⁷+28x⁶+56x⁵+70x⁴ +56x³+28x²+8x+1

Homework:  Construct the 8 the through 10 order of pascals triangle.  I will be posting a large table soon so check back. 

                                                                                                            

¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹   

Binomial Theory Using Polynomial Expansion

If we expand:
 (x+a)(x+b)=x²+(a+b)x+ab
(x+a)(x+b)(x+c)=x³+(a+b+c)x²+(ab+ac+bc)x+abc
(x+a)(x+b)(x+c)(x+d)=x⁴+(a+b+c+d)x³+(ab+ac+ad+bc+bd+cd)x²+(abc+abd +cd+bcd)x+abcd

Lets say that we want to have 2 numbers or a binomial we would change the expressions above to:
(x+1)², (x+1)³, and (x+1)⁴
They equal:
(x+1)²=x²+(1+1)x+1X1=x²+2x+1
(x+1)³=x³+(1+1+1)x²+(1X1+1X1+1X1)x+1X1X1=x³+3x²+3x=1
When when we expand a binomial, the number preceding each x is just the number of combinations of a,b,c variables preceding our x in our polynomial expansion.

(x+1)⁴=x⁴+4x³+6x²+4x+1
(x+1)ⁿ=xⁿ+[nC(n-1)]xⁿ⁻¹+[nC(n-2)]xⁿ⁻²+[nC(n-3)]xⁿ⁻³...........

Now lets lets see what happens if we use a variable like y instead of 1.
(x+y)²=x²+(y+y)x+y.y=x²+2yx+y²
(x+1)³=x³+(y+y+y)x²+(y.y+y.y+y.y)x+y.y.y=x³+3yx²+3y²x+y³
(x+y)⁴=x⁴+4yx³+6x²y²+4xy³+y⁴

Our expansion becomes
(x+y)ⁿ=xⁿ+[nC(n-1)]xⁿ⁻¹y+[nC(n-2)]xⁿ⁻²y²+[nC(n-3)]xⁿ⁻³y³...........yⁿ

Lets say we have (x+y)⁶=x⁶+6x⁵y+15x⁴y²+20x³y³+15x²y⁴+6xy⁵+y⁶

If we had a number like 3 instead of y, we would substitute 3 where we have y and it would look like this:

(x+3)⁶=x⁶+6x⁵(3)+15x⁴(9)+20x³(27)+15x²(81)+6x(243)+y⁶(729)=x⁶+18x⁵+135x⁴+540x³+ 1215x²+1458x+729y⁶

As you can see there is a lot of calculation involved in expanding binomial series.  There is a way that you can just plug a number in from a table. This table is called Pascal's Triangle.  We will do post on this next.  Many people skip binomial expansion and just use Pascal's Triangle.

Polynomial Expansion 

(x+a)(x+b)=x²+(a+b)x+ab
Is the first and many times the last way we learn to expand polynomial equations.  The equation on the left side of the equal sign is known as the factored form and the equation on the right side is known as the general form.  This is very simple.  Third, forth, fifth, and nth order polynomials are very easy to understand also even though most people are not taught how to do it.

Lets look at the expansion of a third order polynomial:

(x+a)(x+b)(x+c)=x³+(a+b+c)x²+(ab+ac+bc)x+abc

And lets look at a forth order expansion:

(x+a)(x+b)(x+c)(x+d)=x⁴+(a+b+c+d)x³+(ab+ac+ad+bc+bd+cd)x²+(abc+abd+acd+bcd)x+abcd

How do we arrive at these expansion solutions?  Lets multiply through and see.

(x+a)(x+b)(x+c)

Lets multiply (x+a)(x+b)=x²+(a+b)x+ab

Lets multiply  (x+a)(x+b)(x+c)=(x+c)(x+a)(x+b)=(x+c)[x²+(a+b)x+ab]

Multiplying x =x³+(a+b)x²+abx
Multiplying c=cx²+c(a+b)x+abc=cx²+(ca+cb)x+abc=

(x+c)[x²+(a+b)x+ab]=x³+(a+b)x²+abx+cx²+(ca+cb)x+abc
Combine terms:  x³+(a+b)x²+cx²+(ca+cb)x+abx+abc=x³+(a+b+c)x²+(ab+ac+bc)x+abc

This is a little tedious but there is a pattern:

• The first term is going to be an exponent of the order of the polynomial.
• The second term is going to be the addition of all the roots times xⁿ⁻¹
• The last term will be the multiplication of all the roots.  This is consistent among all polynomials.
• The middle terms are just permutations of the roots.

So lets look at (x+a)(x+b)(x+c)(x+d)(x+e)(x+f)=

• First lets look at the second term:  That will be (a+b+c+d+e+f)x⁵
• The last term is: abcdef
• The third term is ab+ac+ad+ae+af+bc+bd+be+bf+cd+ce+cf+de+df+ef
• The forth term is abc+abd+abe+abf+acd+ace+acf+ade+adf+aef+bcd+bce+bcf+bde+bdf+bef+cde+cdf +cef+def
• The fifth term: abcd+abce+abcf+acde+acdf+adef+bcde+bcdf+bdef+cdef
• The sixth term is abcde+abcdf+acdef

So (x+a)(x+b)(x+c)(x+d)(x+e)(x+f)=x⁶+(a+b+c+d+e+f)x⁵+(ab+ac+ad+ae+af+bc+bd+be+bf+cd+ce+cf+de+df+ef)x⁴+(abc+abd+abe+abf+acd+ace+acf+ade+adf+aef+bcd+bce+bcf+bde+bdf+bef+cde+cdf

+cef+def )x³+(abcd+abce+abcf+acde+acdf+adef+bcde+bcdf+bdef+cdef)x²+(abcde+abcdf+acdef)x + abcdef

That is pretty long.  The main idea is that is the second term and the last term are always very easy to ascertain.  While we have formulas for the quadratic equation, we do not have similar equations for finding roots of higher order polynomials but there are many things we can use to solve this equation like multiple similar roots, even and odd roots as well as other roots.   We will look at those in the subsequent posts; We will also look at expansion of the equations with the leading or first terms being numbers other than 1;   We will also use these equations to explain the binomial expansion theory; We will also look at algorithms for solving and expanding polynomials.

Lots of good things to follow.

¹  ²  ³  ⁴  ⁵  ⁷  ⁸  ⁹

Fast Multiplication Using Foiling

Many people have used Foiling to distribute quadratic equations.  It actually can be very helpful in multiplying very fast.

When we foil we multiply the first terms together, then we multiply the outside, then the inside, and finally the last terms.  So lets see this in action.

11X15  Not so easy right?

Lets rewrite it:

(10 +1)(10+ 5)

First term is 10X10=100
Outside: 5x10=50
Inside: 1x10=10
Last: 5x1=5
So 100+50+10+5=165

Still doesn't feel to easy?  There are too many numbers so lets modify our foil and lets make three spaces.

100  + 60    +   5     

First       O+I       Last

Okay a little easier

How about these numbers?:

1x1=1

5x1+1x1=6
5x1=5

Quite a bit simpler

Okay line them all up like:        1

                                                   +   6

                                                   +       5

It's 165.

Lets try 13X15:

1x1=                                        1

5x1+3x1=                                     8

3x5 = ?????? what do we do now?

Not to worry.  It is easy.  I promise.

1x1=                                        1

5x1+3x1=                                     8

3x5 =                                             1  5

So our answer is 195.  Since our last multiplier was 2 digits, we just carrier it into the next column which is shared by 8.

How about multiplying a number like (63)(72)?

6x7=                                        4 2

2x6+3x7=                                   33  

3x2 =                                                6

The answer is 4536.  Now that is easy to do on paper but to do it in your head is probably still a little difficult.  Let's break this a part but we are going to start with the middle.  2x6=12 and 3x7=21 they add up to 33.  33 is easy to remember.  We have 6 x 7 is 42 with a carry from 33 of 3 which is 45 and 2 times 3 is 6 so we have 4536.

Okay lets do something even more difficult:  49x83.

Middle is 3x4 +9x8 =12+72=84 remember this number.
Our first number is 8x4=32.
Our last number is 3 times nine 27.
We have a carry from the last digit of 2 and add this to 4=6.
We have a carry of 8 and add that to 32 =40.
So our number is 4067.

Okay lets try:

14x31
Do you visualize and see 434?  If you didn't, I will post more worked process.  It takes a bit of traininng your mind to do these.  After you practice, you will be able to do these really well.

Try these numbers and write down the answer after you figured these out and check with a calculator.
67x85

49x72

88x55

31x96

I will do a post with more worked practice but if you do thes

Foiling Quadratic Functions

(X+A)(X+B) is the multiplication of two factors in a quadratic equation.  We also call it distribution.  A very easy way to solve these equation is with a term called Foiling.  Lets break it down.

First + Outside + Inside + Last

Example:

Expand (X+5)(X+3)

First (X)(X)=X² + Outside (X)(3)=3X +  Inside (5)(X)=5X + Last (5)(3)=15=X² + 8X + 15

Ah ha.  Now you know what foil means.

Lets try expanding (X-4)(X+3)

First (X)(X)=X² + Outside (X)(3)=3X +  Inside (-4)(X)=-4X + Last (-4)(3)=-12=X² -X -12

Something a little harder

Expand (-3X+7)(2X-4)

First (-3X)(2X)=-6X² + Outside (-3X)(-4)=-12X +  Inside (7)(2X)=14X + Last (7)(-24)=-28=

=-6X² +2X -28

Whohoo! Now try a few yourself:

1) (X+5)(X+2)

2) (X-3)(X+5)
3) (X+9)(X+1)

4) (3X+2)(X-7)
5) (6x-4)(3X-2)

Answers (1) x²+7x +10 (2) x²+2x-15 (3) x²+10x+9 (4) 3x²-19x-14 (5) 18x²+8

Converting Decimals to Binary Numbers the Easy Way

In a previous post, I showed what binary numbers are.  We need to understand know how to convert binary numbers to decimal numbers before we proceed.  

There are a few ways to do this but I am going to show you the easiest way.

So to get the idea lets start with number 7.

We know that this will the some of some roots of 2ⁿ.

Lets figure out how to count in binary but before we do that lets see how we count by going up digits  in the decimal system: (this is counting by successive decades instead of units)

1 10 100 1000 10,000 100,000 1,000,000 .............

We count the same way in binary:
1 10 100 1000 10000 10000 1000000

 If we wanted to figure out what the value of a binary digit in the decimal system, it would be as follows:

 1=2⁰ =1 10=2¹=2  100=2²=4 1000= 2³=8

or

n   1 2 3 4  5  6   7   8     9     10    11    12     13     14     15       16       17       18         19           20

n-10 1 2 3 4   5   6   7     8     9     10     11     12     13     14       15       16       17         18          19

2ⁿ  1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 131072 262144 524,288

If you want to make your own list by expanding the series above, just keep multiplying by 2.  This list will allow you to express binary in a decimal value up to 1,000,000 .

Example

 Find the binary equivalent of 7 in the decimal system.  

First we find the largest number on our list less than 7 and that is 4.   4 is represent by 100 in binary

We will subtract 4 from 7:  7-4=3 and we will find the highest number on our list less than or equal to 3 and that is 2 or 10. We will subtract 2 from 3-2=1

So 7 in decimal is 111 in binary.

What did we just do in this example?  We ended up finding the highest digit and then subtracted to find the remainder so we can find the next highest digit and so on.  It is an easy process, but with large numbers, there can be many steps.

Okay lets find the binary equivalent of the decimal 892.
512 is the largest number on our list less than 892.   n-1=9 so n = 10.       892-512=378
256 is the largest number under 378 on our list so n=9. 378-256=122
64 is the largest number under 122 on our list so n=7. 122-64=58
32 is the largest number under 58 so n=6.  58-32=26
16 is the largest number under 26 so n=5.  26-16=10
8 is the largest number under 10 so n=4. 10-8=2
2 is at n=2

Our answer is:  1 1 0 1 1 1 1 0 1 0 or 1101111010
 Place:             10 9 8 7 6 5 4 3 2 1

To learn the concepts pick some large decimal numbers and convert into binary and use a decimal to binary calculator to check your answer.

30 x 30 Table for Squares

Students and also many adults like myself were compelled to learn a multiplication table and to know perfect squares up to 12².  A student who is able to memorize squares up to 30² will be way ahead of other students.  Even learning up to 20² 

Lets see if we can learn up to a 20².  You have a these numbers 169, 196, 225, 256, 289, 324, 361, and 400.  We are going to memorize one number and that is 13²=169.

So we start with 169 and we then reverse the 2 digits and have 196.
We also understand that all numbers that end in 5 always end in 25.  If 14² is 196, then 15² will be larger than 200 so  15²=225.

We take the last 2 digits and move them to the front.  So 225 becomes 25x and 16² has a units digit of 6 so 16²=256.

17² is 289 and the last two digits add up to 17.

324 is 2 3 4 a numerical procession  with 2 and 3 reversed.

361 3+6=9 so if we add them together in our number we have 91.  We flip these around and we have 19.

Another quick way is that the front numbers progress by 3 from 169 3 times and then progress by 4.  The units digit always follows the square of the units digit.
169 196 225 256 289 324 369 400

400 is easy.

There are many people who will contend that there are fast ways to figure out squares and there are but if a person is able to  memorize the table to a 20 x 20 (30 x 30 is even better) they will be incredible mathematicians.  Concepts like long division, factoring polynomials, logarithms, trigonometry, and basic algebra will be easier.  Advanced math like many integration theorems, Greens Theorem  and Fourier Transforms will all be easier if person understands numbers beyond the 10 by 10 multiplication table.  People will also understand and retain this information for life time instead of just forgetting about it when they leave academia.  

	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20	21	22	23	24	25	26	27	28	29	30
1	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20	21	22	23	24	25	26	27	28	29	30
2	2	4	6	8	10	12	14	16	18	20	22	24	26	28	30	32	34	36	38	40	42	44	46	48	50	52	54	56	58	60
3	3	6	9	12	15	18	21	24	27	30	33	36	39	42	45	48	51	54	57	60	63	66	69	72	75	78	81	84	87	90
4	4	8	12	16	20	24	28	32	36	40	44	48	52	56	60	64	68	72	76	80	84	88	92	96	100	104	108	112	116	120
5	5	10	15	20	25	30	35	40	45	50	55	60	65	70	75	80	85	90	95	100	105	110	215	120	125	130	135	140	145	150
6	6	12	18	24	30	36	42	48	54	60	66	72	78	84	90	96	102	108	114	120	126	132	238	144	150	156	162	168	174	180
7	7	14	21	28	35	42	49	56	63	70	77	84	91	98	105	112	119	126	133	140	147	154	261	168	175	182	189	196	203	210
8	8	16	24	32	40	48	56	64	72	80	88	96	104	112	120	128	136	144	152	160	168	176	285	192	200	208	216	224	232	240
9	9	18	27	36	45	54	63	72	81	90	99	108	117	126	135	144	153	162	171	180	189	208	308	216	225	234	243	252	261	270
10	10	20	30	40	50	60	70	80	90	100	110	120	130	140	150	160	170	180	190	200	210	220	230	240	250	260	270	280	290	300
11	11	22	33	44	55	66	77	88	99	110	121	132	143	154	165	176	187	198	209	220	231	242	253	264	275	286	297	308	319	330
12	12	24	36	48	60	72	84	96	108	120	132	144	156	168	180	192	204	216	228	240	252	264	276	288	300	312	324	336	348	360
13	13	26	39	52	65	78	91	104	117	130	143	156	169	182	195	208	221	234	247	260	273	286	299	312	325	338	351	364	377	390
14	14	28	42	56	70	84	98	112	126	140	154	168	182	196	210	224	238	252	266	280	294	308	322	336	350	364	378	392	406	420
15	15	30	45	60	75	90	105	120	135	150	165	180	195	210	225	240	255	270	285	300	315	330	345	360	375	390	405	420	435	450
16	16	32	48	64	80	96	112	128	144	160	176	192	208	224	240	256	272	288	304	320	336	252	368	384	400	416	432	448	464	480
17	17	34	51	68	85	102	119	136	153	170	187	204	221	238	255	272	289	306	323	340	357	274	391	408	425	442	459	476	493	510
18	18	36	54	72	90	108	126	144	162	180	198	216	234	252	270	288	306	324	342	360	378	296	414	432	450	468	486	504	522	540
19	19	38	57	76	95	114	133	152	171	190	209	228	247	266	285	304	323	342	361	380	399	318	437	456	475	492	513	532	551	570
20	20	40	60	80	100	120	140	160	180	200	220	240	260	280	300	320	340	360	380	400	420	440	460	480	500	518	540	560	580	600
21	21	42	63	84	105	126	147	168	189	210	231	252	273	294	315	336	357	378	399	420	441	462	483	504	525	542	567	588	609	630
22	22	44	66	88	110	132	154	176	208	220	242	264	286	308	330	252	274	296	318	440	462	484	506	528	550	572	594	616	638	660
23	23	46	69	92	215	238	261	285	308	230	253	276	299	322	345	368	391	414	437	460	483	506	529	552	575	595	618	641	667	690
24	24	48	72	96	120	144	168	192	216	240	264	288	312	336	360	384	408	432	456	480	504	528	552	576	600	624	648	672	696	720
25	25	50	75	100	125	150	175	200	225	250	275	300	325	350	375	400	425	450	475	500	525	550	575	600	625	650	618	700	725	750
26	26	52	78	104	130	156	182	208	234	260	286	312	338	364	390	416	442	468	494	520	546	572	595	624	650	676	702	728	754	780
27	27	54	81	108	135	162	189	216	243	270	297	324	351	378	405	432	459	486	513	540	567	594	618	648	675	702	729	756	783	810
28	28	56	84	112	140	168	196	224	252	280	308	336	364	392	420	448	476	504	532	560	588	616	641	672	700	728	756	784	812	840
29	29	58	87	116	145	174	203	232	261	290	319	348	377	406	435	464	493	522	551	580	609	638	667	696	725	754	783	812	841	870
30	30	60	90	120	150	180	210	240	270	300	330	360	390	420	450	480	510	540	570	600	630	660	690	720	750	780	810	840	870	900