Hello All,
Help understanding the solve for this one would be much appreciated!
If both 5^2 and 3^3 are factors of n x 2^5 x 6^2 x 7^3, what is the smallest possible positive value of n?
A. 25
B. 27
C. 45
D. 75
E. 125
OA: D
Exponent Factorization
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- Target2009
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Do the prime factorization. So here only 6^2 can be factorize to 2^2 and 3^2.
Ans if 5^2 and 3^3 is factor of given expression then n must have 5^2 * 3^1 = 25 * 3 = 75
So D.
Ans if 5^2 and 3^3 is factor of given expression then n must have 5^2 * 3^1 = 25 * 3 = 75
So D.
Regards
Abhishek
------------------------------
MasterGmat Student
Abhishek
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MasterGmat Student
Thanks for the response Abhishek!
Can you help me understand how you got to n must have a factor of 5^2 and 3^1?
5^2 as a factor is given, is the 3^1 resulting from factoring out 3^2 from 6^2? I think I'm not seeing the relationship here.
Thanks!
Can you help me understand how you got to n must have a factor of 5^2 and 3^1?
5^2 as a factor is given, is the 3^1 resulting from factoring out 3^2 from 6^2? I think I'm not seeing the relationship here.
Thanks!
- Target2009
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For 5^2 and 3^3 to be factor of n x 2^5 x 6^2 x 7^3 you need 5^2 and 3^3 in the expression.Buix0065 wrote:Thanks for the response Abhishek!
Can you help me understand how you got to n must have a factor of 5^2 and 3^1?
5^2 as a factor is given, is the 3^1 resulting from factoring out 3^2 from 6^2? I think I'm not seeing the relationship here.
Thanks!
Now if u prim factorize and see how many 5 and 3 are there you will come to know how many more you need.
n x 2^5 x 6^2 x 7^3 = n x 2^5 x (2^2 x 3^2) x 7^3 = n x 2^7 x 3^2 x 7^3
Now you need two 5 and 1 more 3. so n must be 5^2 * 3^1.
Hope this hepls.
Regards
Abhishek
------------------------------
MasterGmat Student
Abhishek
------------------------------
MasterGmat Student
- tpr-becky
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having a number as a factor also means it could be evenly divided out therefore the statment says:
n x 2^5 x 6^2 x 7^3/(5^2)(3^3) = an integer.
then 6 is not a prime number and so should be factored: 6^2 is the same as 6 x 6 which is the same as
(2x3)(2x3) which is the same as 2^2 x 3^2. put that into the orignal formula:
n x 2^5 x 2^2 x 3^2 x 7^3/ 5^2 3^3 = when we do the division we can cancel (factor out) a 3^2 from the top and bottom. to get
n x 2^5 x 2^2 x 7^3/ 5^2 x 3 - thus in order to factor out the rest of the denominator the n must contain a 5^2 and a 3 - therefore the least possible value of n is 25(3) or 75.
I really pulled this one apart to show technique, once you have mastered the idea of factoring it will come much quicker.
Best of Luck
n x 2^5 x 6^2 x 7^3/(5^2)(3^3) = an integer.
then 6 is not a prime number and so should be factored: 6^2 is the same as 6 x 6 which is the same as
(2x3)(2x3) which is the same as 2^2 x 3^2. put that into the orignal formula:
n x 2^5 x 2^2 x 3^2 x 7^3/ 5^2 3^3 = when we do the division we can cancel (factor out) a 3^2 from the top and bottom. to get
n x 2^5 x 2^2 x 7^3/ 5^2 x 3 - thus in order to factor out the rest of the denominator the n must contain a 5^2 and a 3 - therefore the least possible value of n is 25(3) or 75.
I really pulled this one apart to show technique, once you have mastered the idea of factoring it will come much quicker.
Best of Luck
Becky
Master GMAT Instructor
The Princeton Review
Irvine, CA
Master GMAT Instructor
The Princeton Review
Irvine, CA