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15 Jun 2024, 06:54
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Factor: 6
Not a Factor: 15
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
65% (02:23) correct
35%
(02:14)
wrong
based on 856
sessions
History
Date
Time
Result
Not Attempted Yet
Bartholomew picks a random positive integer greater than 2 and adds 12 to its factorial.
Select for Factor a number that must be a factor of that sum, and select for Not a Factor a number that cannot be a factor of that sum. Make only two selections, one in each column.
Select for Factor a number that must be a factor of that sum, and select for Not a Factor a number that cannot be a factor of that sum. Make only two selections, one in each column.
| Factor | Not a Factor | |
| 4 | ||
| 6 | ||
| 9 | ||
| 12 | ||
| 15 | ||
| 36 |
ShowHide Answer
Official Answer
Factor: 6
Not a Factor: 15
18 Nov 2024, 04:11
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Bunuel
Official Solution:
Factorial of all numbers more than 2 will be divisible by 2 * 3 = 6, so for any \(n\) picked, \(n! + 12\) would represent the sum of two multiples of 6, thus resulting in a multiple of 6. So, 6 must be a factor of \(n! + 12\) regardless of \(n\).
If \(n = 4\), then \(n! + 12 = 36\), so 4, 9 and 12 could also be factors of the sum.
If \(n > 4\), then \(n!\) will be a multiple of 5, so \(n! + 12 = (a \ multiple \ of \ 5) + (not \ a \ multiple \ of \ 5)\), which would not be a multiple of 5. Therefore, for \(n\) equal to 3, 4, or greater than 4, \(n! + 12\) can never be a multiple of 5 and, as a result, can never be a multiple of 15 either.
Therefore, 6 will always be a factor and 15 will never be a factor of \(n! + 12\), where \(n > 2\).
Correct answer:
Factor "6"
Not a Factor "15"
Attachment:
GMAT-Club-Forum-2qjj7qf3.png [ 10.6 KiB | Viewed 2381 times ]
General Discussion
02 Nov 2024, 03:05
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This question appeared on my Mock test and although I got it right during the test, I relied on guesswork for some of the options. During review I found a time-efficient and logical way to solve this question.
Since Bartholomew is picking a factorial greater than 2, he will start with 3! and since he has to pick a value less than 15, he will end at 14!.
So our number becomes x! + 12 where 3<=x<=14
One value that must be a factor is easier so let's get that sorted first. Right from 3! + 12 to 14! + 12, we can take out 6 common from all terms.
3! + 12 => 6(1+2)
4! + 12 => 6(4+2)
...
14! + 12 => 6(4*5*6..*14+2)
So no matter what the value, 6 will always be a factor. Hence, 6 MUST be a factor.
Now to figure out a value that must NOT be a factor, first let's eliminate some answer choices which can be a factor.
I know right off the bat that just like 6, we can have 4! + 12 where we can take 4 common => 4! + 12 => 4(3!+3)
Since 4 can be a factor, it cannot NOT be a factor. Eliminate 4.
Similarly, for 12! + 12, we can take 12 common so 12 can also be a factor. Eliminate 12.
We're now left with 9, 15 and 36. At this point, I'll need to start looking at some values.
let's take some examples and add 12 to them.
3! + 12 => 18
4! + 12 => 36
You can go further, but we can see that 3! + 12 is divisible by 9 and 4! + 12 is divisible by 36 so we can eliminate both 9 and 36 as both of them can be a factor.
The only remaining number is 15. Hence 15 cannot be a factor.
Factor: 6, Not Factor: 15
Since Bartholomew is picking a factorial greater than 2, he will start with 3! and since he has to pick a value less than 15, he will end at 14!.
So our number becomes x! + 12 where 3<=x<=14
One value that must be a factor is easier so let's get that sorted first. Right from 3! + 12 to 14! + 12, we can take out 6 common from all terms.
3! + 12 => 6(1+2)
4! + 12 => 6(4+2)
...
14! + 12 => 6(4*5*6..*14+2)
So no matter what the value, 6 will always be a factor. Hence, 6 MUST be a factor.
Now to figure out a value that must NOT be a factor, first let's eliminate some answer choices which can be a factor.
I know right off the bat that just like 6, we can have 4! + 12 where we can take 4 common => 4! + 12 => 4(3!+3)
Since 4 can be a factor, it cannot NOT be a factor. Eliminate 4.
Similarly, for 12! + 12, we can take 12 common so 12 can also be a factor. Eliminate 12.
We're now left with 9, 15 and 36. At this point, I'll need to start looking at some values.
let's take some examples and add 12 to them.
3! + 12 => 18
4! + 12 => 36
You can go further, but we can see that 3! + 12 is divisible by 9 and 4! + 12 is divisible by 36 so we can eliminate both 9 and 36 as both of them can be a factor.
The only remaining number is 15. Hence 15 cannot be a factor.
Factor: 6, Not Factor: 15
