Class was started off today with another fun problem!

Daily Problem : Find the smallest positive integer which when divided by 10 leaves a remainder of nine, when divided by 9 leaves a remainder of 8, by 8 leaves a remainder of 7, etc…down to where when divided by 2 it leaves a remainder of 1

First suggestions came up that it the number must end in nine, and therefore also must be odd. This is because if the number was even there is a possibility of not having a remainder, and if it was any other odd number, there would be more than one for the remainder.

Whenever we see “remainder” we have to think modulo.

x=9 (mod 10)          so x+1 ? ( mod 10) which is congruent to 0 (mod 10)

continuing this pattern…

x=8 (mod8)               so x+1 ? 0(mod9)

x=7 (mod8)              x+1 ? 0(mod8)

continuing…

x=1 (mod2)                 x+1 ? (mod2)

x+1 must be evenly divisible by 9, 8, 7, … 1

now think that 10 requires a 2 and a 5

9 requires a 3 and a 3

6 requires a 2 and a 3

4 requires a 2 and a 2

2 requires a 2 and a 1

somehow if you multiple 2 5 3 3 2 2 7 then subtract 1, it gives you the answer

this can also be written as x+1= (2^3) (3^2) (5) (7) which equals 2,519

the problem is, i am not yet familiar on how this answer came about.  I will ask Mr. Bieniek, and get back to this lovely post asap. sorry for the inconvenience but i do not want to give the inter-web incorrect information.

Today’s actual lesson consisted of reviewing awesome transformations.

the first problem was what happens when .5 f(x) is graphed.  one must remember that this is an output function and therefore does not affect the scale, but changes the graph.  Also because it is an output function all of the X values will remain the same but the Y values will change accordingly.

the second problem is now f(.5 x)  This time it is an input function.  Because it is input the scale must change.  If you graph it on another grid with the correct scale, it appears to be the same graph.  but if you regraph it using the original scale but the new x values, the y values remain the same as the x values are either stretched or scrunched.

f(x)-4  What will happen now.  Well this one is simple, the entire graph is moved down four spaces.

f(x-7)  this one is different than the previous funciton. this is an input function so the scale must be changed. Because it is f(x-7) the entire grid is pulled 7 unites to the left (because it’s negative). Therefore the new graph appears to have moved to the right. 7 becomes lies on the original origin and other than being moved on the x-axis, the graph remains the same.

And the next lucky contestant to have the honor of posting a post to this blog is……….(drum role please)………………. MISS SHANNA ****!!!!!

After lunch, we walk down the hall with our stomachs full, conversation in the air, and a smile on our face… that is until we enter math class and see the problem on the board:

Given  9^6=531441 , how would you represent 531440 in base 9?

Our math system is in base 10. if you look at the number: 5 4 6 9 1 we can break it down in which the base of 10 is represented.

1 we know is in the “ones” place, but what that really means is 1 is in the 10^0 in which we all know 10^0 is 1. likewise 9 is in the “tens” place meaning 10^1 which of course equals 10. and we can continue down the line in which 6 is in the 10^2 (hundreds), 4 is in the 10^3 (thousands), and 5 is in the 10^4 (ten thousands). and the line continues with however big the number is which is why our math system is in the base 10.

Now, when working in the base of ten we only use the numbers 0-9 in each place. To best understand this look at a odometer. lets say there is a total of 145368 miles. you travel another mile thus bringing the odometer to 145369, but once you go up another mile you will not change the ones place to 10 it will drop back down to 0 and add 1 digit to the tens place.

If we must alter the base and convert it to base 9 (as stated in the problem) how would we go about that? well we know the base of 10 uses the numbers 0-9, thus we know the base of 9 will use the numbers 0-8. lets refer back to the problem:

Given 9^6=531441, how would you represent 531440 in base 9.

We know 9^6 = 531441, but we must get 1 less than that 531440.

Looking back in terms of base 10, lets say we need 1 less than 10000 what would that be. well of course it is 9999. Likewise one less than 531441 in terms of base 9 will produce 5 digits. because we know 9^6 produces exactly 531441, we now can find that 88888 = 531440 in base 9. The ones place there is an 8 (obviously), in base 9 the ones place is represented by 9^0. likewise you continue with that same concept til you reach the ten thousands place which is represented by 9^5 in base 9.

To check your answer multiply each digit by base 9 represented in that certain place value and add them all together:

8*9^5=472392 +

8*9^4=52488    +

8*9^3=5832       +

8*9^2=648          +

8*9^1=72             +

8*9^0=8               +

all equals 531440 in base 10.

If your still having problems converting bases try these practice problems:

Convert 43 in base 5 to base 10.

(4*5^1=20 + 3*5^0= 3 = 23 in base 10)

Convert 56 in base 4 to base 10.

Convert 93 in base 3 to base 9.

If you need further explanation visit this website:

http://www.learn-programming.za.net/articles_decbinhexoct.html

tomorrows blogger will be Steph

Today, I walked into precalculus, sat down, and looked at the smart board. The word “confused” did not even cover how i felt about the problem that we had to try to solve. The question was:

What is the remainder when 7^348 +25^605 is divided by eight?

After sitting without a clue of what to do, Mr. B finally explained a method called modular arthimetic. He said it was almost like how a clock works. The clock goes up to 12 but then goes back to 1 again and so on and so forth. Modular arithmatic is basically doing math in cycles. In this case, since we are divided the sum by 8, we have to solve the problem using cycles of 8 or modulo 8 [mod 8].

First, we will start with 7^248. Since 8 can’t go into 7 any times, we have to use a higher exponent of 7. So 7^2 is 49. You then have to divide 49 by eight and see what your remainder is. 8 goes into 49 six times (48) with a remainder of 1. Therefore, 7^2 is congruent to 1 in mod 8. Now to find out what exactly 7^348 is, we can use 7^2 to find the remainder. Since, 7^2 is congruent to 1 in mod 8, then we can replace it, in the below equation. (Don’t forget that an exponent of an exponent can be multiplied and and exponent can be divided to make an exponent of an exponent. In the following problem, 2 times 174 equal the originial exponent of 348.)

7^348 = (7^2)^174 is congruent to 1^174 [mod 8] is congruent to 1 [mod 8].

The remainder of 7^348 is 1.

Now, moving on to 25^605, since 8 can go into 25, it isn’t nessecary to use any exponents to find the remainder. Because goes 8 into 25 three times (24), the remainder is 1. Therefore, 25 is congruent to 1 (mod 8). So now we can replace 25 in the below equation.

25^605 is congruent to 1^605 [mod 8] which is congruent to 1 [mod 8]. 

The remainder of 25^605 is 1.

We’re not quite done yet though. Since the expression was additive, we must add the remainders of the two numbers together to get the remainder of their sum. So you take the 1 from the 7^348 and the 1 from the 25^605, and you have a total remainder of 2.

The remainder of 7^348 + 25^605 is 2.

If you are still having difficulties understanding, goes to the following link. This websites explains this method by using amounts of time for examples.

http://www.sci.brooklyn.cuny.edu/~gurwitz/fipse/modtxt.html

Your scribe for tomorrow will be Katie L.

Today, the class was in the computer lab. So far this week, we’ve been working with translations and dilation of  functions. We’ve learned that:

g(x)= a * f(x) affects the outputs of the functions, producing a horizontal dilation. This is caused by the scale of the x axis being changed by the factor of ‘a’. Say a=2. Each square that used to equal one, now equals 2. In comparison to the original function, g(x) would be 1/2 the size. **The factor that the graph is dilated by is the reciprocal of a.

g(x)= f(x) +/- b also affects the outputs of the graph, causing a vertical translation. This leaves the imputs where they are, and moves the function up or down the x axis.

g(x)= f(cx) affects the imputs of the function, stretching or squishing the function in comparison to the preimage, which is a vertical dilation. This happens because the scale on the y axis is changing. For example, if c=1/2, each square that used to equal 1, now equals 1/2. The graph would be stretched to twice the size. **The factor that the graph is dilated by is  the reciprocal of c.

g(x)= f(x +/- d) also affects the imputs of the function, causing a horizontal translation. When looking at the graph, it seems the function itself moves, though actually, it’s the grid that is being shifted left or right. The outputs themselves are not moving.

We took this knowledge, and took it to the next level. The activity we completed in the computer lab dealt with reflections of functions. We were working in Geometer’s Sketchpad, a program that allows you to plot functions, trace points, and track patterns among other features.

We began by reflecting the same point across the x and y axis. Point P is in Quadrant I and is the point being reflected. Point Q is in Quadrant II, reflected across the y axis. Point R is in Quadrant IV, reflected across the x axis. Then, we moved the graph and traced the points. That means, wherever we moved the graph to, a trail was left showing where the point had been before.

Observations showed that the trails of all three points matched exactly. From there we plotted the function f(x)= 1 + sqrt(x-1) and merged point p onto the function. By animating the point, and documenting some coordinates, a relationship was found between the three points.

If point P = (x,y), then point Q = (-x,y) **Q was reflected across the y axis

If point P = (x,y), then point R = (x, -y) **R was reflected across the x axis

This relationship along with the information we already knew (stated above) allowed us to come up with the equations for the reflected functions. My equation for point Q was g(x)= f(-x) and my equation for point R was h(x)= f(-x). The former stated dealt with changing the imputs of the function, which meant the negative sign would need to be placed inside the parenthesis. The latter mentioned affected the outputs of the function, which is why the negative sign was placed outside the function. For further review on the reasons behind, see above.

Unfortunately, that was as much as I completed of the activity before the bell rang. If you wish for more explanation in a different format on this topic, see

http://www.analyzemath.com/graphs_functions/graph_reflections.html

http://www.themathpage.com/aprecalc/reflections.htm

The next scribe will be Alyssa.