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5 + 3 n is NEVER

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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
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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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by sadullaevd » Fri Feb 19, 2010 5:09 am
Cant find any perfect square,

so it could be perfect square, choice E
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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



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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,
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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.
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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 :)



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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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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 :)
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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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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?
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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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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
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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?

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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.. :(
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