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 n is NEVER
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Puttingsanju09 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
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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Cant find any perfect square,
so it could be perfect square, choice E
so it could be perfect square, choice E
Stay skeptical,
Think critically,
Assume nothing.
Think critically,
Assume nothing.
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-1 as well as 1 are both 1 in magnitudeharsh.champ wrote:Puttingsanju09 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
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]
The mind is everything. What you think you become. -Lord Buddha
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And can anyone explain what "magnitude" is?
What do we need it for?
Thanks in advance,
What do we need it for?
Thanks in advance,
Stay skeptical,
Think critically,
Assume nothing.
Think critically,
Assume nothing.
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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.sadullaevd wrote:And can anyone explain what "magnitude" is?
What do we need it for?
Thanks in advance,
The mind is everything. What you think you become. -Lord Buddha
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Ohh,fell into the question trap.sanju09 wrote:-1 as well as 1 are both 1 in magnitudeharsh.champ wrote:Puttingsanju09 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
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]
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.
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A 5+6 =11 is primesanju09 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
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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5 + 3 * 2 = primesanju09 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 * 3 = divisible by 7
5 + 3 * (-2) = 1 in magnitude
5 + 3 * 1 = perfect cube
by POE : E
Regards,
Harsha
Harsha
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If 5 + 3 n were the perfect square of some integer, say p, thenshashank.ism wrote:A 5+6 =11 is primesanju09 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
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..
5 + 3 n = p^2
3 n = p^2 - 5
But, p^2 - 5 is NEVER divisible by 3. WHY?
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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?sanju09 wrote:If 5 + 3 n were the perfect square of some integer, say p, thenshashank.ism wrote:A 5+6 =11 is primesanju09 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
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..
5 + 3 n = p^2
3 n = p^2 - 5
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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Becausesanju09 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?
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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Isn't that p^2 - 5 = ( p^2 - 1) - 4?ajith wrote:Becausesanju09 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?
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
Rest is fabulous!!
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I was about to edit the post.. you caught me before that..sanju09 wrote: Isn't that p^2 - 5 = ( p^2 - 1) - 4?
Rest is fabulous!!
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