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08 Jan 2025, 01:01
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If \(n\) is an integer, which of the following could be a factor of both \(4n - 1\) and \(8n + 40\)?
I. 7
II. 21
III. 42
A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III
I. 7
II. 21
III. 42
A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III
08 Jan 2025, 01:01
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Official Solution:
If \(n\) is an integer, which of the following could be a factor of both \(4n - 1\) and \(8n + 40\)?
I. 7
II. 21
III. 42
A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III
First, notice that since \(n\) is an integer, \(4n - 1\) will always be odd. This means it cannot have any even factors, so 42 is eliminated from the start.
Next, let's rewrite \(8n + 40\) so that \(4n - 1\) appears in the expression. This could help us deduce a connection between the two numbers:
\(8n + 40 =\)
\(= (8n - 2) + 42 =\)
\(= 2(4n - 1) + 42\).
Now, we need to find a divisor, \(d\), such that both \(\frac{4n - 1}{d}\) and \(\frac{2(4n - 1) + 42}{d}\) are integers. Observe that:
\(\frac{2(4n - 1) + 42}{d}= \frac{2(4n - 1)}{d} + \frac{42}{d}\).
Now, if \(\frac{4n - 1}{d}\) is an integer, then \(\frac{2(4n - 1)}{d}\) will also be an integer. This means that for \(\frac{2(4n - 1)}{d} + \frac{42}{d}\) to be an integer, \(\frac{42}{d}\) must also be an integer. Consequently, \(d\) must be an odd factor of 42.
The odd factors of 42 are 1, 3, 7, and 21 (and their negatives).
Answer: C
If \(n\) is an integer, which of the following could be a factor of both \(4n - 1\) and \(8n + 40\)?
I. 7
II. 21
III. 42
A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III
First, notice that since \(n\) is an integer, \(4n - 1\) will always be odd. This means it cannot have any even factors, so 42 is eliminated from the start.
Next, let's rewrite \(8n + 40\) so that \(4n - 1\) appears in the expression. This could help us deduce a connection between the two numbers:
\(8n + 40 =\)
\(= (8n - 2) + 42 =\)
\(= 2(4n - 1) + 42\).
Now, we need to find a divisor, \(d\), such that both \(\frac{4n - 1}{d}\) and \(\frac{2(4n - 1) + 42}{d}\) are integers. Observe that:
\(\frac{2(4n - 1) + 42}{d}= \frac{2(4n - 1)}{d} + \frac{42}{d}\).
Now, if \(\frac{4n - 1}{d}\) is an integer, then \(\frac{2(4n - 1)}{d}\) will also be an integer. This means that for \(\frac{2(4n - 1)}{d} + \frac{42}{d}\) to be an integer, \(\frac{42}{d}\) must also be an integer. Consequently, \(d\) must be an odd factor of 42.
The odd factors of 42 are 1, 3, 7, and 21 (and their negatives).
Answer: C
09 Apr 2025, 04:28
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Hi Bunuel ,
42 can also be a factor right since 42/d where d = 42 is also an integer , in that case the answer choice should be E.
42 can also be a factor right since 42/d where d = 42 is also an integer , in that case the answer choice should be E.
Bunuel
