5 + 3 n is NEVER

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

by sanju09 » Fri Feb 19, 2010 4:40 am

For some integer n, 5 + 3 n is NEVER
(A) a prime
(B) divisible by 7
(C) 1 in magnitude
(D) a perfect cube
(E) a perfect square

The mind is everything. What you think you become. -Lord Buddha

Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com

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Source: Beat The GMAT — Problem Solving | Fri Feb 19, 2010 4:40 am

harsh.champ: Legendary Member; Posts: 1132; Joined: Mon Jul 20, 2009 3:38 amFollowed by: 6 members; Location: India; Thanked: 64 times; GMAT Score:760

by harsh.champ » Fri Feb 19, 2010 5:01 am

sanju09 wrote:For some integer n, 5 + 3 n is NEVER
(A) a prime
(B) divisible by 7
(C) 1 in magnitude
(D) a perfect cube
(E) a perfect square

Putting
n=1 we get 8 which is a perfect cube.
n=2 we get 11 which is a prime.
n=3 we get 14 which is divisible by 7

[spoiler]It will never be 1 in magnitude since n has to be -4/3 which is not an integer.Hence,C.[/spoiler]

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sadullaevd: Senior | Next Rank: 100 Posts; Posts: 81; Joined: Wed Apr 08, 2009 3:12 pm; Location: South Korea; Thanked: 4 times

by sadullaevd » Fri Feb 19, 2010 5:09 am

Cant find any perfect square,

so it could be perfect square, choice E

Stay skeptical,
Think critically,
Assume nothing.

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

by sanju09 » Fri Feb 19, 2010 5:09 am

harsh.champ wrote:
sanju09 wrote:For some integer n, 5 + 3 n is NEVER
(A) a prime
(B) divisible by 7
(C) 1 in magnitude
(D) a perfect cube
(E) a perfect square
Putting
n=1 we get 8 which is a perfect cube.
n=2 we get 11 which is a prime.
n=3 we get 14 which is divisible by 7

[spoiler]It will never be 1 in magnitude since n has to be -4/3 which is not an integer.Hence,C.[/spoiler]

-1 as well as 1 are both 1 in magnitude

The mind is everything. What you think you become. -Lord Buddha

Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com

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sadullaevd: Senior | Next Rank: 100 Posts; Posts: 81; Joined: Wed Apr 08, 2009 3:12 pm; Location: South Korea; Thanked: 4 times

by sadullaevd » Fri Feb 19, 2010 5:11 am

And can anyone explain what "magnitude" is?

What do we need it for?

Thanks in advance,

Stay skeptical,
Think critically,
Assume nothing.

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

by sanju09 » Fri Feb 19, 2010 5:23 am

sadullaevd wrote:And can anyone explain what "magnitude" is?

What do we need it for?

Thanks in advance,

Magnitude of a number on the real number line is the distance (a scalar quantity) of the number from 0 of the number line. It's same as the modulus or absolute value of any number.

The mind is everything. What you think you become. -Lord Buddha

Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com

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harsh.champ: Legendary Member; Posts: 1132; Joined: Mon Jul 20, 2009 3:38 amFollowed by: 6 members; Location: India; Thanked: 64 times; GMAT Score:760

by harsh.champ » Fri Feb 19, 2010 5:32 am

sanju09 wrote:
harsh.champ wrote:
sanju09 wrote:For some integer n, 5 + 3 n is NEVER
(A) a prime
(B) divisible by 7
(C) 1 in magnitude
(D) a perfect cube
(E) a perfect square
Putting
n=1 we get 8 which is a perfect cube.
n=2 we get 11 which is a prime.
n=3 we get 14 which is divisible by 7

[spoiler]It will never be 1 in magnitude since n has to be -4/3 which is not an integer.Hence,C.[/spoiler]
-1 as well as 1 are both 1 in magnitude

Ohh,fell into the question trap.
For n=-2 we get -1 which is 1 in magnitude.
Hence answer would be E.
Thanks sanju09 for pointing out this error.On the test day,I would have surely marked C as the option.

It takes time and effort to explain, so if my comment helped you please press Thanks button

Just because something is hard doesn't mean you shouldn't try,it means you should just try harder.

"Keep Walking" - Johnny Walker

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shashank.ism: Legendary Member; Posts: 1022; Joined: Mon Jul 20, 2009 11:49 pmFollowed by: 2 members; Location: Gandhinagar; Thanked: 41 times

by shashank.ism » Fri Feb 19, 2010 7:46 am

sanju09 wrote:For some integer n, 5 + 3 n is NEVER
(A) a prime
(B) divisible by 7
(C) 1 in magnitude
(D) a perfect cube
(E) a perfect square

A 5+6 =11 is prime
B 5+ 9 = 14 divisble by 7
C 5-6 = -1 mag = 1
D 5+3 = 8 a perfect cube
E left option is E so I go with E is there a way to check E..

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harshavardhanc: Legendary Member; Posts: 526; Joined: Sat Feb 21, 2009 11:47 pm; Location: India; Thanked: 68 times; GMAT Score:680

by harshavardhanc » Fri Feb 19, 2010 11:32 am

sanju09 wrote:For some integer n, 5 + 3 n is NEVER
(A) a prime
(B) divisible by 7
(C) 1 in magnitude
(D) a perfect cube
(E) a perfect square

5 + 3 * 2 = prime

5 + 3 * 3 = divisible by 7

5 + 3 * (-2) = 1 in magnitude

5 + 3 * 1 = perfect cube

by POE : E

Regards,
Harsha

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outreach: Legendary Member; Posts: 748; Joined: Sun Jan 31, 2010 7:54 amFollowed by: 3 members; Thanked: 46 times

by outreach » Fri Feb 19, 2010 11:51 am

answer is E

a) for n =-1 cond satisfied
b) for n=3 cond satisfied
c) for n =-2 cond satisfied
d) for n=1 cond satisfied

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

by sanju09 » Sat Feb 20, 2010 12:44 am

shashank.ism wrote:
sanju09 wrote:For some integer n, 5 + 3 n is NEVER
(A) a prime
(B) divisible by 7
(C) 1 in magnitude
(D) a perfect cube
(E) a perfect square
A 5+6 =11 is prime
B 5+ 9 = 14 divisble by 7
C 5-6 = -1 mag = 1
D 5+3 = 8 a perfect cube
E left option is E so I go with E is there a way to check E..

If 5 + 3 n were the perfect square of some integer, say p, then

5 + 3 n = p^2

3 n = p^2 - 5

But, p^2 - 5 is NEVER divisible by 3. WHY?

The mind is everything. What you think you become. -Lord Buddha

Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com

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shashank.ism: Legendary Member; Posts: 1022; Joined: Mon Jul 20, 2009 11:49 pmFollowed by: 2 members; Location: Gandhinagar; Thanked: 41 times

by shashank.ism » Sat Feb 20, 2010 12:58 am

sanju09 wrote:
shashank.ism wrote:
sanju09 wrote:For some integer n, 5 + 3 n is NEVER
(A) a prime
(B) divisible by 7
(C) 1 in magnitude
(D) a perfect cube
(E) a perfect square
A 5+6 =11 is prime
B 5+ 9 = 14 divisble by 7
C 5-6 = -1 mag = 1
D 5+3 = 8 a perfect cube
E left option is E so I go with E is there a way to check E..
If 5 + 3 n were the perfect square of some integer, say p, then

5 + 3 n = p^2

3 n = p^2 - 5

But, p^2 - 5 is NEVER divisible by 3. WHY?

yeah sanju u have tried a good thing ...but now also I m not able to go beyond this ...But, p^2 - 5 is NEVER divisible by 3. WHY?
Have you solved it further ..could u or anyone else here explain this...though we have reached to our solution..but we need to know.....Instructors please help...

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ajith: Legendary Member; Posts: 1275; Joined: Thu Sep 21, 2006 11:13 pmFollowed by: 2 members; Location: Arabian Sea; Thanked: 125 times

by ajith » Sat Feb 20, 2010 2:06 am

sanju09 wrote:
If 5 + 3 n were the perfect square of some integer, say p, then

5 + 3 n = p^2

3 n = p^2 - 5

But, p^2 - 5 is NEVER divisible by 3. WHY?

Because

p^2-5=( p^2-1) +4
= ( p+1) (p-1) +4

Now p-1, p, p+1 are three consecutive integers , one of them should be a multiple of 3

If p-1 or p+1 is a multiple of 3; p^2-5 is not because 4 is not a multiple of 3
if p is a multiple of 3 p^2 is a multiple of 3 but p^2 -5 is not a multiple of 3

Always borrow money from a pessimist, he doesn't expect to be paid back.

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

by sanju09 » Sat Feb 20, 2010 2:21 am

ajith wrote:
sanju09 wrote:
If 5 + 3 n were the perfect square of some integer, say p, then

5 + 3 n = p^2

3 n = p^2 - 5

But, p^2 - 5 is NEVER divisible by 3. WHY?
Because

p^2-5=( p^2-1) +4
= ( p+1) (p-1) +4

Now p-1, p, p+1 are three consecutive integers , one of them should be a multiple of 3

If p-1 or p+1 is a multiple of 3; p^2-5 is not because 4 is not a multiple of 3
if p is a multiple of 3 p^2 is a multiple of 3 but p^2 -5 is not a multiple of 3

Isn't that p^2 - 5 = ( p^2 - 1) - 4?

Rest is fabulous!!

The mind is everything. What you think you become. -Lord Buddha

Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com

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ajith: Legendary Member; Posts: 1275; Joined: Thu Sep 21, 2006 11:13 pmFollowed by: 2 members; Location: Arabian Sea; Thanked: 125 times

by ajith » Sat Feb 20, 2010 2:23 am

sanju09 wrote: Isn't that p^2 - 5 = ( p^2 - 1) - 4?

Rest is fabulous!!

I was about to edit the post.. you caught me before that..

Always borrow money from a pessimist, he doesn't expect to be paid back.

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