dabaobao wrote:
GMATPrepNow wrote:
gmatt1476 wrote:
In an auditorium, 360 chairs are to be set up in a rectangular arrangement with x rows of exactly y chairs each. If the only other restriction is that 10 < x < 25, how many different rectangular arrangements are possible?
A. Four
B. Five
C. Six
D. Eight
E. Nine
PS66602.01
From the given information, the TOTAL number of chairs = xy
This means: xy = 360
Since x and y must be POSITIVE INTEGERS, there is a finite number of possibilities.
To help us list the pairs of values with a product of 360, let's find the prime factorization of 360
360 = (2)(2)(2)(3)(3)(5)
When we consider the fact that 10 < x < 25, the possibilities are:
x = 12 & y = 30
x = 15 & y = 24
x = 18 & y = 20
x = 20 & y = 18
x = 24 & y = 15
There are five such possibilities
Answer: B
Cheers,
Brent
ScottTargetTestPrep This question is simple but took a bit of time in order to avoid missing any values of x. While I know how to solve it instantly by taking factor pairs, it did take me 3 min for me to double check to make sure I didn't miss any value of x in that range. Is there a quick to find all values of x that satisfy the condition 10 < x < 25 since we can have (3+1)(2+1)(1+1) = 24 values of x? Thanks!
Since you know there are (3+1)(2+1)(1+1) = 24 possible values of x, I understand you already factored 360 = 2^3 * 3^2 * 5.
First of all, you definitely don't need to consider all 24 possibilities for x since we are only interested in the values of x that satisfy 10 < x < 25. In order to make sure we are not missing any factor pairs, we will find all values of x in the interval 10 < x < 25 that are a factor of 360.
We already have the prime factorization 360 = 2^3 * 3^2 * 5; therefore we can immediately eliminate the primes in the interval 10 < x < 25 which do not appear in the prime factorization of 360. Namely, we eliminate the values x = 11, 13, 17, 19 and 23. Next, we can eliminate any multiples of 7 and 11 as well (since 360 cannot be divisible by any multiple of 7 or 11). Thus, the values x = 14, x = 21 and x = 22 are also eliminated. We are left with x = 12, 15, 16, 18, 20, 24. At this point, we can simply check these six values and determine that 360 is divisible by all except 16 (too few factors of 2 in 360 to be divisible by 16). This is the fastest way I can come up with which will guarantee that you won't miss any factor pairs.
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