Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

If n is a two-digit positive integer, what is the probability that n^4 -5n^2+4 is not a multiple of 20?

(A) 1/10 (B) 1/6 (C) 1/5 (D) 5/6 (E) 9/10

C. n^4 -5n^2+4 = (n-2) (n-1) (n+1)(n+2) if n is a multiple of 5, then only it is not divisible by 20. so the prob = 1/5. very good question:

I disagree. 2^5 = 32 will fit the bill. There is more to this. let me continue working.

After working for a few more mins., I found the answer as 1/5. Now hold on. We are said that n is a 2 digit number. so we have 90 numbers to consider and there are 17 2 digit multiples of 5. As I said 32 will also fit the bill. So the answer is 18/90 = 1/5.

Last edited by dahcrap on 23 Aug 2007, 08:32, edited 1 time in total.

I see your point...can you show me how you worked it out???

dahcrap wrote:

Himalayan wrote:

kevincan wrote:

If n is a two-digit positive integer, what is the probability that n^4 -5n^2+4 is not a multiple of 20?

(A) 1/10 (B) 1/6 (C) 1/5 (D) 5/6 (E) 9/10

C. n^4 -5n^2+4 = (n-2) (n-1) (n+1)(n+2) if n is a multiple of 5, then only it is not divisible by 20. so the prob = 1/5. very good question:

I disagree. 2^5 = 32 will fit the bill. There is more to this. let me continue working.

After working for a few more mins., I found the answer as 1/5. Now hold on. We are said that n is a 2 digit number. so we have 90 numbers to consider and there are 17 2 digit multiples of 5. As I said 32 will also fit the bill. So the answer is 18/90 = 1/5.

I see your point...can you show me how you worked it out???

dahcrap wrote:

Himalayan wrote:

kevincan wrote:

If n is a two-digit positive integer, what is the probability that n^4 -5n^2+4 is not a multiple of 20?

(A) 1/10 (B) 1/6 (C) 1/5 (D) 5/6 (E) 9/10

C. n^4 -5n^2+4 = (n-2) (n-1) (n+1)(n+2) if n is a multiple of 5, then only it is not divisible by 20. so the prob = 1/5. very good question:

I disagree. 2^5 = 32 will fit the bill. There is more to this. let me continue working.

After working for a few more mins., I found the answer as 1/5. Now hold on. We are said that n is a 2 digit number. so we have 90 numbers to consider and there are 17 2 digit multiples of 5. As I said 32 will also fit the bill. So the answer is 18/90 = 1/5.

From the stem, I deduced there are 90 numbers to consider. Then I got the equation (n-2)(n+2)(n-1)(n+1). My instinct told me to first consider powers of numbers 2 to 9 that are 2 digit numbers. This is where I guess I lost 2 valuable mins doing so. I found 2^5 and 5^2. But then there was no answer as 1/45. I had to find atleast 9 numbers to get the least of the answer choices 1/10. Then it hit me. Why not start with 5. I found 5 along with 25 satisfied the equation but had to eliminate 5 because it was a 1 digit number. Then I considered a few more multiples of 5 and I felt happy that I could reach 1/5. I spent a few more mins checking some more multiple but found none. Over all I would have spent abt 7-8 mins doing this problem. On second thoughts, i cud have checked all the 90 numbers manually

If n is a two-digit positive integer, what is the probability that n^4 -5n^2+4 is not a multiple of 20?

(A) 1/10 (B) 1/6 (C) 1/5 (D) 5/6 (E) 9/10

C. n^4 -5n^2+4 = (n-2) (n-1) (n+1)(n+2) if n is a multiple of 5, then only it is not divisible by 20. so the prob = 1/5. very good question:

I disagree. 2^5 = 32 will fit the bill. There is more to this. let me continue working.

After working for a few more mins., I found the answer as 1/5. Now hold on. We are said that n is a 2 digit number. so we have 90 numbers to consider and there are 17 2 digit multiples of 5. As I said 32 will also fit the bill. So the answer is 18/90 = 1/5.

for what?

i didnot say 2^5, i said if n is a multiple of 5. for ex: 10, 15, 20 .........95. and there are 18, not 17, multiples of 5

if n = 10, (n-2) (n-1) (n+1)(n+2) = 8x9x11x12, which is not divisible by 20.
if n = 11, (n-2) (n-1) (n+1)(n+2) = 9x10x12x13, which is divisible by 20.
if n = 12, (n-2) (n-1) (n+1)(n+2) = 10x11x13x14, which is divisible by 20.
if n = 13, (n-2) (n-1) (n+1)(n+2) = 11x12x14x15, which is divisible by 20.
if n = 14, (n-2) (n-1) (n+1)(n+2) = 12x13x15x16, which is divisible by 20.

keep on doing, 1 out of 5 is not divisible by 20 if n is a multiple of 5.

yeah but suppose if N=64 which is not a multiple of 5, in that case

n-2=62 n-1=63 n+1=65 n+2=66

62*63*65*66 is a multiple of 20.....

suppose n=61

59*60*62*63 is a multiple of 20...

61-64=3 so here we have 2 examples where you have 2 multiples of 20 3 numbers apart...

I thought i had this figured out....now i am all confused...

Himalayan wrote:

dahcrap wrote:

Himalayan wrote:

kevincan wrote:

If n is a two-digit positive integer, what is the probability that n^4 -5n^2+4 is not a multiple of 20?

(A) 1/10 (B) 1/6 (C) 1/5 (D) 5/6 (E) 9/10

C. n^4 -5n^2+4 = (n-2) (n-1) (n+1)(n+2) if n is a multiple of 5, then only it is not divisible by 20. so the prob = 1/5. very good question:

I disagree. 2^5 = 32 will fit the bill. There is more to this. let me continue working.

After working for a few more mins., I found the answer as 1/5. Now hold on. We are said that n is a 2 digit number. so we have 90 numbers to consider and there are 17 2 digit multiples of 5. As I said 32 will also fit the bill. So the answer is 18/90 = 1/5.

for what?

i didnot say 2^5, i said if n is a multiple of 5. for ex: 10, 15, 20 .........95. and there are 18, not 17, multiples of 5

if n = 10, (n-2) (n-1) (n+1)(n+2) = 8x9x11x12, which is not divisible by 20. if n = 11, (n-2) (n-1) (n+1)(n+2) = 9x10x12x13, which is divisible by 20. if n = 12, (n-2) (n-1) (n+1)(n+2) = 10x11x13x14, which is divisible by 20. if n = 13, (n-2) (n-1) (n+1)(n+2) = 11x12x14x15, which is divisible by 20. if n = 14, (n-2) (n-1) (n+1)(n+2) = 12x13x15x16, which is divisible by 20.

keep on doing, 1 out of 5 is not divisible by 20 if n is a multiple of 5.

dahcrap is really crappy..he got me to forget what the question was...grrr

Hayabusa wrote:

fresinha12 wrote:

yeah but suppose if N=64 which is not a multiple of 5, in that case

n-2=62 n-1=63 n+1=65 n+2=66

62*63*65*66 is a multiple of 20.....

suppose n=61

59*60*62*63 is a multiple of 20...

61-64=3 so here we have 2 examples where you have 2 multiples of 20 3 numbers apart...

I thought i had this figured out....now i am all confused...

Himalayan wrote:

dahcrap wrote:

Himalayan wrote:

kevincan wrote:

If n is a two-digit positive integer, what is the probability that n^4 -5n^2+4 is not a multiple of 20?

(A) 1/10 (B) 1/6 (C) 1/5 (D) 5/6 (E) 9/10

C. n^4 -5n^2+4 = (n-2) (n-1) (n+1)(n+2) if n is a multiple of 5, then only it is not divisible by 20. so the prob = 1/5. very good question:

I disagree. 2^5 = 32 will fit the bill. There is more to this. let me continue working.

After working for a few more mins., I found the answer as 1/5. Now hold on. We are said that n is a 2 digit number. so we have 90 numbers to consider and there are 17 2 digit multiples of 5. As I said 32 will also fit the bill. So the answer is 18/90 = 1/5.

for what?

i didnot say 2^5, i said if n is a multiple of 5. for ex: 10, 15, 20 .........95. and there are 18, not 17, multiples of 5

if n = 10, (n-2) (n-1) (n+1)(n+2) = 8x9x11x12, which is not divisible by 20. if n = 11, (n-2) (n-1) (n+1)(n+2) = 9x10x12x13, which is divisible by 20. if n = 12, (n-2) (n-1) (n+1)(n+2) = 10x11x13x14, which is divisible by 20. if n = 13, (n-2) (n-1) (n+1)(n+2) = 11x12x14x15, which is divisible by 20. if n = 14, (n-2) (n-1) (n+1)(n+2) = 12x13x15x16, which is divisible by 20.

keep on doing, 1 out of 5 is not divisible by 20 if n is a multiple of 5.

so where do you disagree? i did not get you.

Read the problem again.

gmatclubot

Re: DS: Probability
[#permalink]
23 Aug 2007, 12:34