Math questions for you.

[thirtythree]

I have some questions from a math test that I couldn't figure out. Would someone explain them to me? BTW, we weren't allowed to use a calculator.

[1] How many four-digit numbers are there that meet these requirements?
1. Only use digits 1,3,4,6,8, and 9
2. Are between 3300 and 7200
3. Are even
4. Have no repeating digits

[2] What is the inverse function of: y=3ln[x+(x^2+1)^(1/2)]

[3] (I think I got this one) An equilateral triangle is inscribed in a circle and another circle is enscribed in that triangle. What is the ratio of the circles' areas?

I'll post some others if I can remember them.

 

[thelanx]

1. 84 I think
2. I know how to do it...but too much algebra...and I don't have pencil and paper here...
3. 1:4

If you want explanations...tell me.

 

[thirtythree]

Originally posted by: thelanx
1. 84 I think
2. I know how to do it...but too much algebra...and I don't have pencil and paper here...
3. 1:4

If you want explanations...tell me.

Thanks! Could you explain number 1?

 

[Hanpan]

1. THis is a combinatorics question. YOu need to choose 4 numbers from 1,3,4,6,8,9 (6numbers) with replacement and order is important.
So you have 6 choices for the first, 6 for the second 6 for the third and 6 for the fourth. Therefore you have 6^4 choices or 1296 numbers that contain 4 digits.
2. This is the same idea expect the 3 is fixed, you have only (3,4,6,8,9) 5 numbers to choose from for the other 3 digits. Then repeat for 4 fixed plus 3more digits, and 6 fixed. Add together results for each.
3. first 3 digits are any of the 6 numbers and the fourth digit is one of 4,6,8.
4. Now order is important. So you have 6 numbers to pick from first, then 5, then 4 etc.

Here is a blurb on this kind of idea.

From ask Dr. Math

Sorry don't have time to do the others right now...

EDIT> Seems I cannot count. There are only 6 nubmers not seven. Thereore take every number I used and decrement it by one.

Edit2 changed to reflect this

 

[thirtythree]

Originally posted by: Hanpan
1. THis is a combinatorics question. YOu need to choose 4 numbers from 1,3,4,6,8,9 (7numbers) with replacement and order is important.
So you have 7 choices for the first, 7 for the second 7 for the third and 7 for the fourth. Therefore you have 7^4 choices or 2401 numbers that contain 4 digits.
2. This is the same idea expect the 3 is fixed ...

Wow, now I am really . I don't think we have reached the .. uhh, combinatorics section in my pre-calc class.

 

[blahblah99]

Originally posted by: Deslocke
I have some questions from a math test that I couldn't figure out. Would someone explain them to me? BTW, we weren't allowed to use a calculator.

[1] How many four-digit numbers are there that meet these requirements?
1. Only use digits 1,3,4,6,8, and 9
2. Are between 3300 and 7200
3. Are even
4. Have no repeating digits

[2] What is the inverse function of: y=3ln[x+(x^2+1)^(1/2)]

[3] (I think I got this one) An equilateral triangle is inscribed in a circle and another circle is enscribed in that triangle. What is the ratio of the circles' areas?

I'll post some others if I can remember them.

#1 and #4. With no other restrictions, you have a 6-digit number. You have 6 choices in selecting the first digit, 5 choices in selecting 2nd digit, etc etc.... so you have basically 6*5*4*3*2*1 = 6!

#2. With a restriction of being 4 digits AND between 3300-7200, you are limited to 3 numbers for first digit, 4, 6, 8, and 9 if you pick 3 as first digit, 1, 3, 6, 8, and 9 if you pick 4 as first digit, and 1, 3, 4, 8, and 9 if you pick 6 as first digit, etc etc...

#3. Since there are 3 even digits, 3 odd digits, even numbers are just half of what you come up with for #2.

 

[Hanpan]

Ok.

Look at it this way.

YOu have 6 things. (numbers) For a four digit number you need to get four of these 6 things.

So for your first digt you can choose any of the 6 things (numbers)
For your second digit you can again choose any of the 6.
Same for 3 and 4.

This gives you 6 choices for digit 1 times 6 choices for digit 2 times 6 choices for digit 3 times 6 choices for digit four = 6*6*6*6 = 6^4

For the second part.

If you want a number between 3300 and 7200, we have to ensure the first digit does not go above above 7. This involves two parts.

First, we keep the first digit 3 constant. Now if we want above 3300 we cannot have a 1 in the seoncd place so we must choose from only the remaining 5 numbers. Finally for the 3rd and fourth digits we can choose any of the 6 numbers again. This gives us 1*5*6*6.
Second, we take the numbers greater than 3999 but less than 7200. This allows us only two possible numbers for the first digit, 4 or 6. The remaining 3 can be any of the 6 so we have 2*6*6*6. we now add the two results for the answer.

For the third part. If the number is to be even you can choose any of the 6 for the first 3 digits but the last digit must be one of 4,6,8 for the number to be even. Following the earlier trend we have 6*6*6*3. Notice this is the result for the first part divided by 2.

Finally for the fourth part you now have 6 numbers for the first digit, but since you cannot repeat you now only have 5 numbers left to choose the second digit from. The third must be chosen from the remaining 4, and finally the fourht from the remaining 3. This gives us 6*5*4*3

Again I apologize for not being able to count and using 7 in my original example. There were only 6 numbers.

I hope this is a little more clear. If not shoot me off a pm.