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# If n is a two-digit positive integer, what is the

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If n is a two-digit positive integer, what is the [#permalink]  23 Aug 2007, 06:04
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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
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Re: DS: Probability [#permalink]  23 Aug 2007, 06:18
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:
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OK here is my approach

i break down the equation to (n-2)(n-1)N(n+1)(n+2)

question says N is two digit number..we are looking for the equation to have at least 2 2s and 1 5.

Now if N is a multiple of 5..then (n-2)(n-1)(n+1)(n+2) will not have a multiple of 5 in it..and thus not a multiple of 20!

lets see all multiples of 5...from 10-95=18 Ok!

now we have 100-10 =90 2 digit numbers... so total numbers =90

18/90=2/10=1/5

1/5 it is..
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"question says N is two digit number..we are looking for the equation to have at least 2 2s and 1 5.

Now if N is a multiple of 5..then (n-2)(n-1)(n+1)(n+2) will not have a multiple of 5 in it..and thus not a multiple of 20!"

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Re: DS: Probability [#permalink]  23 Aug 2007, 08:25
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.

Last edited by dahcrap on 23 Aug 2007, 08:32, edited 1 time in total.
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Re: DS: Probability [#permalink]  23 Aug 2007, 08:52
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.
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Re: DS: Probability [#permalink]  23 Aug 2007, 09:15
fresinha12 wrote:
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
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Re: DS: Probability [#permalink]  23 Aug 2007, 09:33
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.
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Re: DS: Probability [#permalink]  23 Aug 2007, 12:02
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.
Current Student
Joined: 28 Dec 2004
Posts: 3391
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Re: DS: Probability [#permalink]  23 Aug 2007, 12:34
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.

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