• We’re currently investigating an issue related to the forum theme and styling that is impacting page layout and visual formatting. The problem has been identified, and we are actively working on a resolution. There is no impact to user data or functionality, this is strictly a front-end display issue. We’ll post an update once the fix has been deployed. Thanks for your patience while we get this sorted.

Math Puzzles *UPDATED*

Page 3 - Seeking answers? Join the AnandTech community: where nearly half-a-million members share solutions and discuss the latest tech.
Originally posted by: Vertimus
Originally posted by: whitecloak
1/4

series 1 : 1/2 +1/4 + 1/8 + ...
series 2 : 1/5 + 1/4 + 5/16 + ...
Click to expand...

Nope. Your second series doesn't converge to 1.
Click to expand...

it seems to converge.

1/5 + (1/5)/(4/5) + (1/5)/((4/5)*(4/5)) + ...= (1/5)/[1-(4/5)] = 1
 
Originally posted by: whitecloak
Originally posted by: Vertimus
Originally posted by: whitecloak
1/4

series 1 : 1/2 +1/4 + 1/8 + ...
series 2 : 1/5 + 1/4 + 5/16 + ...
Click to expand...

Nope. Your second series doesn't converge to 1.
Click to expand...

it seems to converge.

1/5 + (1/5)/(4/5) + (1/5)/((4/5)*(4/5)) + ...= (1/5)/[1-(4/5)] = 1
Click to expand...

bah. I suck. found out the mistake.

 
Originally posted by: Vertimus
Originally posted by: chuckywang
9) (sqrt(5)-1)/8
Click to expand...

Can you provide a short explanation please?
Click to expand...

Hah, I would think with an answer that random that if it were right, I knew what I was doing.

Anyways, if 1/8 is the third term and r is the ratio, then 1/(8r^2) is the first term. Therefore, 1/(8*r^2)/(1-r) = 1. This simplifies to a cubic equation, and you know that one of the roots is r=1/2 since 1/2+1/4+1/8+...is one of those sequences. The other two roots can be found using the quadratic formula. Only one of them is positive and less than 1, that being r=(1+sqrt(5))/4, so the second term is 1/8*4/(1+sqrt(5)) = (sqrt(5)-1)/8
 
Originally posted by: chuckywang
Originally posted by: Vertimus
Originally posted by: chuckywang
9) (sqrt(5)-1)/8
Click to expand...

Can you provide a short explanation please?
Click to expand...

Hah, I would think with an answer that random that if it were right, I knew what I was doing.

Anyways, if 1/8 is the third term and r is the ratio, then 1/(8r^2) is the first term. Therefore, 1/(8*r^2)/(1-r) = 1. This simplifies to a cubic equation, and you know that one of the roots is r=1/2 since 1/2+1/4+1/8+...is one of those sequences. The other two roots can be found using the quadratic formula. Only one of them is positive and less than 1, that being r=(1+sqrt(5))/4, so the second term is 1/8*4/(1+sqrt(5)) = (sqrt(5)-1)/8
Click to expand...

Yep, good job! :beer:
 
Originally posted by: Vertimus
Originally posted by: chuckywang
10) 54
Click to expand...

I need a explanation before I can accept any solution. Please read the first post.
Click to expand...

This one is harder to explain, but I'll try to explain the main points.
First of all 2004 = 2^2*3*167 is the prime factorization.
Note that for the prime factorization of any number: p1^a1*p2^a2*p3^a3*...., there are (1+a1)*(1+a2)*(1+a3)*.... differenct divisors of that number.

Therefore, if you want a divisor of 2004 to have 2004 different divisors, then those numbers must have the prime factorization of:

2^a1*3^a2*167^a3

and must it must be true that (1+a1)*(1+a2)*(1+a3) = 2004.

Basically, how many different ways can you multiply three positive numbers to get 2004, and ORDER MATTERS.

Running through them all, you can count 54 different ways.
 
Actually, it basically was counting.. You may call it cheating but I just wrote a simple script to do the work for me. Mathematically speaking, meh.

Edit: BTW, that was my 2^7th post. 😀
 
Originally posted by: Vertimus
Originally posted by: whitecloak
Originally posted by: Vertimus
Originally posted by: whitecloak
1359?
Click to expand...

Explanation? For which problem?
Click to expand...

oops. for problem 11.

remove numbers which have 3 different digits and 4 digits from 9999.
Click to expand...

I think you need to explain more 😛
Click to expand...


Number of possible 3 digit numbers with 3 different digits = 10*9*8
Number of possible 4 digit numbers with 3 different digits = 10*9*8*4 (4-takes care of the ordering)
Number of possible 4 digit numbers with 4 different digits = 10*9*8*7
Total numbers with more than 2 diff digits = 8640
Total numbers with less than or equal 2 diff digits = 9999 -8640 = 1359
 
Originally posted by: whitecloak
Originally posted by: Vertimus
Originally posted by: whitecloak
Originally posted by: Vertimus
Originally posted by: whitecloak
1359?
Click to expand...

Explanation? For which problem?
Click to expand...

oops. for problem 11.

remove numbers which have 3 different digits and 4 digits from 9999.
Click to expand...

I think you need to explain more 😛
Click to expand...


Number of possible 3 digit numbers with 3 different digits = 10*9*8
Number of possible 4 digit numbers with 3 different digits = 10*9*8*4 (4-takes care of the ordering)
Number of possible 4 digit numbers with 4 different digits = 10*9*8*7
Total numbers with more than 2 diff digits = 8640
Total numbers with less than or equal 2 diff digits = 9999 -8640 = 1359
Click to expand...

1359 is incorrect. I don't have time to say why it's wrong, since i'm off to dinner.
 
You must log in or register to reply here.

TRENDING THREADS

Back
Top