Is the positive integer n equal to the square of an integer?
(1) For every prime number p, if p is a divisor of n, then so is p2.
(2) sqrt root (n) is an integer.
I got this correct. But I'm still not sure about Stmt 1. Pls. Explain. Thks!
Prime & Square
-
- Master | Next Rank: 500 Posts
- Posts: 125
- Joined: Mon Dec 15, 2008 9:24 pm
-
- Master | Next Rank: 500 Posts
- Posts: 197
- Joined: Sun May 18, 2008 2:47 am
- Thanked: 12 times
joyseychow wrote:Is the positive integer n equal to the square of an integer?
(1) For every prime number p, if p is a divisor of n, then so is p2.
(2) sqrt root (n) is an integer.
I got this correct. But I'm still not sure about Stmt 1. Pls. Explain. Thks!
statement 1)
restate:
for every prime number that is a factor of n then p^2 is also a factor of n.
let's make our prime number(p) = 5. N at least would have to equal 25 in order for p^2 to be a factor of N. So n=p^2. However, n could have additional primes that are not p. For instance n could equal 50, whose factors are 5^2 and 2. If this were the case n does not equal the square of an integer. It equals the square of an integer times another prime number. So insufficient.
Statement 2)
sqroot n = an integer
n = integer squared
sufficient
B is answer.
pls confirm.
you got this!
-
- Master | Next Rank: 500 Posts
- Posts: 103
- Joined: Mon May 04, 2009 11:53 am
- Thanked: 6 times
My answer is D.
A.) I am translating this to mean that if there is “any” prime number in N, then the p^2 version of that prime is also in N. Meaning – Every prime is squared.
If this is the correct interpretation, then this is sufficient. A perfect square, when reduced to primes, will have even numbers as exponents of those primes. For Example.
2*3*5*7 = Not a perfect square
2^2*3^2*5^2*7^2 = perfect square.
Sufficient.
A.) I am translating this to mean that if there is “any” prime number in N, then the p^2 version of that prime is also in N. Meaning – Every prime is squared.
If this is the correct interpretation, then this is sufficient. A perfect square, when reduced to primes, will have even numbers as exponents of those primes. For Example.
2*3*5*7 = Not a perfect square
2^2*3^2*5^2*7^2 = perfect square.
Sufficient.
Last edited by navalpike on Fri Aug 07, 2009 9:04 am, edited 1 time in total.
I think the answer is D as well.
If a number is a perfect square, all of its prime factors have to occur an even number of times in its factorization. Which means, if the number is divisible by a prime number n, it should be divisble by n^2 as well.
So if the corollary holds good as well.
Example -
Take 100 which is 10^2
(5*2)^2 - so the number is divisible by 5 and 5^2 and 2 and 2^2.
OA please?
If a number is a perfect square, all of its prime factors have to occur an even number of times in its factorization. Which means, if the number is divisible by a prime number n, it should be divisble by n^2 as well.
So if the corollary holds good as well.
Example -
Take 100 which is 10^2
(5*2)^2 - so the number is divisible by 5 and 5^2 and 2 and 2^2.
OA please?
-
- Master | Next Rank: 500 Posts
- Posts: 197
- Joined: Sun May 18, 2008 2:47 am
- Thanked: 12 times
you may be right. my interpretation is wrong. we need OAs...zeenab wrote:I think the answer is D as well.
If a number is a perfect square, all of its prime factors have to occur an even number of times in its factorization. Which means, if the number is divisible by a prime number n, it should be divisble by n^2 as well.
So if the corollary holds good as well.
Example -
Take 100 which is 10^2
(5*2)^2 - so the number is divisible by 5 and 5^2 and 2 and 2^2.
OA please?
-
- Legendary Member
- Posts: 1161
- Joined: Mon May 12, 2008 2:52 am
- Location: Sydney
- Thanked: 23 times
- Followed by:1 members
Answer shouldn't be D...
Statement 1 -
n = 45, p = 3
p is a divisor of n
also p^2 is a divisor of n
but n in this case is not a square of an integer.
Statement 1 -
n = 45, p = 3
p is a divisor of n
also p^2 is a divisor of n
but n in this case is not a square of an integer.
shahdevine wrote:you may be right. my interpretation is wrong. we need OAs...zeenab wrote:I think the answer is D as well.
If a number is a perfect square, all of its prime factors have to occur an even number of times in its factorization. Which means, if the number is divisible by a prime number n, it should be divisble by n^2 as well.
So if the corollary holds good as well.
Example -
Take 100 which is 10^2
(5*2)^2 - so the number is divisible by 5 and 5^2 and 2 and 2^2.
OA please?
-
- Legendary Member
- Posts: 752
- Joined: Sun May 17, 2009 11:04 pm
- Location: Tokyo
- Thanked: 81 times
- GMAT Score:680
edited..joyseychow wrote:Is the positive integer n equal to the square of an integer?
(1) For every prime number p, if p is a divisor of n, then so is p2.
(2) sqrt root (n) is an integer.
I got this correct. But I'm still not sure about Stmt 1. Pls. Explain. Thks!
IMO B- look for scoobydooby's example
1) it says that all the prime factors of n have even powers- i was wrong
it says that the powers are greater than 2. Can be even. Can be odd
Not sufficient
2) Sufficient
@mehravikas,
1) says every prime number.
In your example 45= 3^2*5.
you haven't considered 5.
Last edited by tohellandback on Thu Aug 06, 2009 10:57 pm, edited 1 time in total.
The powers of two are bloody impolite!!
-
- Legendary Member
- Posts: 1035
- Joined: Wed Aug 27, 2008 10:56 pm
- Thanked: 104 times
- Followed by:1 members
would go with B
stmnt 1 is not sufficient by itself.
if N=9, divisible by 3 and 3^2 . yes N perfect square
if N=27, divisble by 3 and 3^2. no N not perfect square
stmnt 2 sufficient
all perfect squares have integers as their square roots.
stmnt 1 is not sufficient by itself.
if N=9, divisible by 3 and 3^2 . yes N perfect square
if N=27, divisble by 3 and 3^2. no N not perfect square
stmnt 2 sufficient
all perfect squares have integers as their square roots.
-
- Master | Next Rank: 500 Posts
- Posts: 197
- Joined: Sun May 18, 2008 2:47 am
- Thanked: 12 times
cool i'm not crazy...for a minute i was racking my brains. I thought I was from another planet when everyone was getting Ds instead of B.scoobydooby wrote:would go with B
stmnt 1 is not sufficient by itself.
if N=9, divisible by 3 and 3^2 . yes N perfect square
if N=27, divisble by 3 and 3^2. no N not perfect square
stmnt 2 sufficient
all perfect squares have integers as their square roots.
-
- Master | Next Rank: 500 Posts
- Posts: 125
- Joined: Mon Dec 15, 2008 9:24 pm
-
- Legendary Member
- Posts: 621
- Joined: Wed Apr 09, 2008 7:13 pm
- Thanked: 33 times
- Followed by:4 members
The question is a bit confusing.. esp stmt 1.
But this is a good discussion.
I got B with my interpretation.
thanks everyone.
-V
But this is a good discussion.
I got B with my interpretation.
thanks everyone.
-V
-
- Master | Next Rank: 500 Posts
- Posts: 208
- Joined: Sat Jan 31, 2009 11:32 am
- Location: Mumbai
- Thanked: 2 times
Is the positive integer n equal to the square of an integer?
(1) For every prime number p, if p is a divisor of n, then so is p2.
(2) sqrt root (n) is an integer.
D
(1) For every prime number p, if p is a divisor of n, then so is p2.
(2) sqrt root (n) is an integer.
D
-
- Senior | Next Rank: 100 Posts
- Posts: 43
- Joined: Thu Mar 12, 2009 8:56 pm
- Thanked: 10 times
- Followed by:1 members
- GMAT Score:730
B
You can test statement 1 by setting n equal to 9 or 18 and p equal to 3.
If n = 9 and p = 3, 9/3 and 9/9 yield integers.
If n = 18 and p = 3, 18/3 and 18/9 yield integers.
However, 18 is NOT a square of an integer. Statement 1 is thus insufficient.
You can test statement 1 by setting n equal to 9 or 18 and p equal to 3.
If n = 9 and p = 3, 9/3 and 9/9 yield integers.
If n = 18 and p = 3, 18/3 and 18/9 yield integers.
However, 18 is NOT a square of an integer. Statement 1 is thus insufficient.