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12 Nov 2024, 02:18
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Factor: 6
Not a Factor: 15
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 |
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Official Answer
Factor: 6
Not a Factor: 15
12 Nov 2024, 02:18
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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"
Bunuel
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"
18 Nov 2024, 04:09
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I have revised the question, solution, and formatting by adding more details to enhance clarity
