Sam2304! superb example given by you... I simply loved that. You gave an example and that is the only way to give an answer as x is an integer greater than 1...
Thank You once again...
x=the 12th power of an integer?
- [email protected]
- Legendary Member
- Posts: 934
- Joined: Tue Nov 09, 2010 5:16 am
- Location: AAMCHI MUMBAI LOCAL
- Thanked: 63 times
- Followed by:14 members
IT IS TIME TO BEAT THE GMAT
LEARNING, APPLICATION AND TIMING IS THE FACT OF GMAT AND LIFE AS WELL... KEEP PLAYING!!!
Whenever you feel that my post really helped you to learn something new, please press on the 'THANK' button.
LEARNING, APPLICATION AND TIMING IS THE FACT OF GMAT AND LIFE AS WELL... KEEP PLAYING!!!
Whenever you feel that my post really helped you to learn something new, please press on the 'THANK' button.
- meanjonathan
- Senior | Next Rank: 100 Posts
- Posts: 42
- Joined: Mon Jan 10, 2011 6:57 pm
- Thanked: 6 times
Shantanu, I love this explanation. You start with the statements and works toward the stem.shantanu86 wrote:
x = a^3
taking 4th power of both sides..
x^4 = a^12 ... (1.)
x= b^4
taking 3rd power of both sides..
x^3 = b^12 ... (2.)
Now, dividing (1.) by (2.) we have-
x = (a/b)^12 ..
Here, a/b has to be an integer hence [C] is the correct choice.
While some of the other explanations offer still a valuable intuitive approach, there is a downside--they start with the stem (assuming its plausibility) and work toward the statements. Though this works in this example, it can be dangerous in general.
To expound, if we start with, for instance, 2^1, work up to 2^6 = 64, and then square that, the result is indeed an example of an integer raised to the 12 power. On it's face, this seems to satisfy the sufficiency requirement.
(2^6)^2= 2^12=4096.
This proves the existence of at least one integer equal a 3rd, 4th, and 12th power; however, the question isn't asking whether its possible for this to occur, it's asking whether the specific number x does and whether we can prove that based on the statements. Therefore, our logic must start at the statements and work toward the stem.
In a sense, the above approach starts with the answer a^12 and works backward to prove that a^12=b^4=c^3. In this case, we luck out--all a^12s equal b^4s and c^3s. But using this approach as your standard for DS will sometimes lead you astray as you're more likely to miss opportunities to prove a statement insufficient. This is especially true with inequalities.
Alternatively, Shantanu starts with the statements and works toward the stem. This way he proves, not just the possibility, but the rule.
Artfully done.
So just to recap--
1) x=(a*a*a) --> x^4 = (a*a*a)(a*a*a)(a*a*a)(a*a*a)
2) x=(b*b*b*b) --> x^3 = (b*b*b*b)(b*b*b*b)(b*b*b*b)
Both Together) (x^4)/(x^3) = (a^12)/(b^12)
Here's the kicker:
Since(x^4)/(x^3) = x^(4-3) = x^1
x =(a^12)/(b^12) = (a/b)^12
So x=(a/b)^12 where a is divisible by b, and therefore an integer. In a sense, this becomes a divisibility problem.
--mj.
-
- Master | Next Rank: 500 Posts
- Posts: 126
- Joined: Sun Jun 24, 2012 10:11 am
- Location: Chicago, IL
- Thanked: 36 times
- Followed by:7 members
There is a very easy explanation why both statements together are sufficient.
(1) x is a cube of integer, means that the prime factorization of x contains primes raised to powers that are multiples of 3 like 0, 3, 6, 9, etc.
(2) x is a 4th power of an integer, means that the prime factorization of x contains primes raised to powers that are multiples of 4
(1) and (2) taken together imply that the prime factorization of x contains primes at powers that are BOTH mulitples of 4 and multiples of 3, which means the powers are multiples of 12 (since 3 and 4 are not divisible by each other).
If the prime factorization of a number contains primes to powers multiples of 12, it means x is the 12th power of an integer.
(1) x is a cube of integer, means that the prime factorization of x contains primes raised to powers that are multiples of 3 like 0, 3, 6, 9, etc.
(2) x is a 4th power of an integer, means that the prime factorization of x contains primes raised to powers that are multiples of 4
(1) and (2) taken together imply that the prime factorization of x contains primes at powers that are BOTH mulitples of 4 and multiples of 3, which means the powers are multiples of 12 (since 3 and 4 are not divisible by each other).
If the prime factorization of a number contains primes to powers multiples of 12, it means x is the 12th power of an integer.
Skype / Chicago quant tutor in GMAT / GRE
https://gmat.tutorchicago.org/
https://gmat.tutorchicago.org/
- GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3