A few more questions

Date: Jan 9, 2005

A few more questions

langris5 · Senior Member

the first one...i know how to do it, but i don't know how to explain it..but i am certain the webmaster can explain it to you in detail when he gets here, just be patient...but i'll show you wut i did...

77/7=11(remeber that)

7+77=84

84/2=42

11*42=462

63. The last digit of 9^54 is

uh...this one is easy, all you have to do is divide the exponent by 4 and find the remainder, then take the base number to the "remainderth power" and find the last digit

so for this one...9^54,

54/4=remainder is 2

9^2=81, the last digit is 1

67. If x is in the first quadrant, and sin x is 4/5, what is sin(2x)?

remember that this is a special 3,4,5, right triangle so sin x is 4/5, therefore cos x = 3/5

and sin(2x) = 2sinxcosx = 2(4/5)(3/5)=24/25

the answer is 24/25

69. If N/9 has a remainder of 7, then (N^4)/9 has a remainder of ______

take the remainder to the 4th power and find the remainder.

so in this case, 7^4/9=2401/9, the remainder is 7

70*. 343089 / 30.7 / 27

i would've missed this one...cuz i'd do 343089/(30x30) and that'll be out of the range...

quincy

langris5 · Senior Member

i was just practicing ns and the 1st test i did dr.numsen 25, march,2004 medium, and i saw all the questions u asked on there, just wondering if you got ur questions from that test....lol

quincy

webmaster · Administrator

Vinay,

Quincy is quite right about the first one. Whenever you are adding a sequesnce in which all the terms differ by the same number, you multiply the number of terms by the first plus the last all divided by 2. In other words:

(first+last)*(number of terms)/2

In this example it becomes: (77+7)*11/2 = 42*11 = 462

-----------------------------------------

Now for the second question. Again Quincy is on the money, but I will try to explain why it is the way it is.

Whenever I see these I always try to find out how many different LAST digits there are. You have to evaluate for each digit until it repeats. Let's look at 7 for example (I will only wirte down the last digits).

7^0 = 1
7^1 = 7
7^2 = 9
7^3 = 3
7^4 = 1 (since we get a number that is repeating we have only 4 unique last digits, so in this case we would need to use n MOD 4, or the remainder when n is divided by 4 in 7^n.

So for example,

7^42 has a last digit of ____. 42 MOD 4 = 2. So this is the same units digit as 7^2 or 9.

7^215 has a last digit of ___. 215 MOD 4 = 3. So this is the same units digit as 7^3 or 3.

In your example you would use MOD 4 again by the same reasoning as above and follow Quincy's instructions.

----------------------------

For you third question, AGAIN good job Quincy. sin(2x) = 2sinxcosx so in this example we are given sin(x) = 4/5. Remember sin(x) = opposite/hypotenuse. So we have this:

adjacent = ?
opposite = 4
hypotenuse = 5

We also know that cos(x) = adjacent / hypotenuse

So sin(x) = 4/5, what does cos(A) equal? It is 3/5 since this is a right triangle.

Try some more that are similar:

1. If sec(x) = -13/5, then sin(x) = _____
2. If tan(x) = 7/24, then cos(x) = ______
3. If cos(x) = 3/5, then cos(2x) = _______

--------------------------------------

I promise I will get to your other 2 questions as soon as I can, but I have to go to work now.

Hope this helped.

Webmaster

-- Edited by webmaster at 08:54, 2005-01-10

webmaster · Administrator

Now for the other two questions:

n/9 has a remainder of 7, then n^4/9 has a remainder of ___________

Well, on this one I will take only a SLIGHTLY different approach than Quincy. It is true that you can use just the remainder on this one, and it is true you can evaluate 7^4 fairly easily to be 2401, but that is not necessary. On this one I would split 7^4 up into managable parts. 7^4 = 7^2 x 7^2 = 49 x 49. So when we look at 49 x 49 divided by 9 we can take each separately and get 49 MOD 9 x 49 MOD 9 = (4 x 4) MOD 9 = 16 MOD 9 = 7. Now 7 is the correct answer. I think this is easier than evaluating 7^4, because this evaluation will not always be easy.

Try these:

1. n/4 has a remainder of 3, then n^12 has a remainder of ______

2. n/6 has a remainder of 3, then n^5 has a remainder of _______

3. n/9 has a remainder of 5, then n^6 has a remainder of _______

--------------------------------------------------------------

Now for the last one, an approximation:

70*. 343089 / 30.7 / 27 = _______

Anytime you are dividing by two numbers like you are in this example you can always change it to be:

343089/(30.7 x 27) which is approximately 343000/810 which is about 3400/8 which is 425. This is close enough in the range. This is how I would do the problem.

*One word of CAUTION: It is okay to change the numerator and play with it, but be very careful when changing the denominator. Try to keep it as close to what it actually is as possible. You can drop the .7 but don't change the 27 to 30. That is a significant change. I was able to successfully go from 810 to 800 ONLY by dropping the numerator by 3000 and I could have just changed it to 3200/8 and still be in the range. If you change the denominator a significant amount, you have to change the numerator by a significant amount as well.

The ONLY way to get good at these is to do LOTS of practice on them. Try making up your own problems and seeing how close you can get to the correct answer.

Let me know if there are any other problems I can help you out on!

Webmaster

-- Edited by webmaster at 16:56, 2005-01-10

Date: Jan 10, 2005

Thanks a lot,

I really understand the questions now. Yeah, I did get the questions from the mar. 2004 test. I've been taking those tests everyday to get better.

For the questions that Mr. Jones posted, here are my answers:

1. N^12/4 will have a remainder of 1

I think that's the answer, but checking it on the calculator, I get zero. How does that work?

2. 3

3. 1

Are these right? I raised the remainder to the power and divided by the original dividend.

Thanks again,

Vinay

By the way, what did you score on that test, quincy? Am I going to have to compete against you this year? lol

langris5 · Senior Member

i figured there would be an easier way to do this problem, thanks!

quincy

langris5 · Senior Member

uh..i don't remember, i usually do 5 or 6 medium tests a day, and i've been working on column 4 problems lately, so i don't really keep track of my scores, i may end up competing against you this year, it depends on if you are in 5A or not and what school you're going to, it also depends on if i can make it to state or not...i have A&M in my district and all the Kleins in my region, but now i am not as worried as i used to be, because now i really begin to enjoy the fun i gain when i compete with time and myself (of course i still wanna win), but anyways how large is ur school and what district are you in?

good luck

quincy

Date: Jan 10, 2005

I'm in 5A

But I'm a freshman. Will we still compete?

Vinay

langris5 · Senior Member

yeah, in UIL we'll compete, it doesn't matter wut grade you are, what school are you going to and how far can you get on those tests?

quincy

Date: Jan 10, 2005

I didn't see the other problems.

Trig

1. -5/13

2. 24/25

3. -9/25 ?

Date: Jan 10, 2005

Um, on the Dr. Numsen UIL level tests, I usually get to 60-70

On the medium, usually 50-60

My scores are usually 260+ on UIL

and 190+ on Medium

is that good enough?

vinay

langris5 · Senior Member

yeah that's awesome for a freshman, just keep workin and u'll get better and better, btw check your trig answers, got i some different solutions

quincy

Date: Jan 10, 2005

hmm...

if sec x is -13/5, then it has to be 5 12 13, with 5 being adjacent, so sin x has to be - oh, ok - 12/13

I think number 2 and 3 are right

thanks

vinay

Date: Jan 10, 2005

btw,

how hard are the regional tests compared to dr. numsen? what score should you get to advance?

vinay

langris5 · Senior Member

....very hard...*unpleasant memories* Mr.White usually makes regional and state tests very very very very.....*different* and hard, i took last year's regional test few wks ago and scored bad. if you are interested...print one from texasmath.org and see how you score

quincy

webmaster · Administrator

Vinay,

You are right, n^12/3 will have a remainder of 1, if n/4 has a remainder of 3.

Keep working and asking if you have separate problems.

Webmaster....

3^12 = 531441

The other questions you did get right

-- Edited by webmaster at 08:45, 2005-01-11

Date: Jan 12, 2005

One more question...

How would you do

11. 11^81 / 8 has a remainder of __________?

I think you would do 3^81 divided by 8, but there has to be an easier way

please let me know,

Vinay

webmaster · Administrator

Vinay,

You are right on the target, you just need to keep going. Check it out.

First, break it up into managable parts:

You could do (3^3)^27, but that still leaves you in a little bit of a pickle. We know that 3^2/8 has a remainder of 1. So make it (3^2)^40 x 3. Now it is super easy:

(3^2)^40 x 3 / 8 has a remainder equal to 9^40 MOD 8 x 3 MOD 8 = 1 x 3 = 3.

3 is your answer.

Webmaster

Date: Jan 14, 2005

Thanks,

I took the regional test that you mentioned and scored a 209. Just curious, but how does this compare to your score last year? Also, we had a UIL test today in school (don't know if it was practice, my teacher said something about a mail-in competition and I scored 265.

Vinay