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21 May 2024, 01:30
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Question Stats:
77% (01:40) correct
23%
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wrong
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If the least common multiple of integers \(\frac{n}{2}\) and \(\frac{n}{3}\) is 210, how many distinct prime factors does the positive integer \(n\) have?
A. 2
B. 3
C. 4
D. 5
E. 6
A. 2
B. 3
C. 4
D. 5
E. 6
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02 Jun 2024, 16:30
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Official Solution:
If the least common multiple of integers \(\frac{n}{2}\) and \(\frac{n}{3}\) is 210, how many distinct prime factors does the positive integer \(n\) have?
A. 2
B. 3
C. 4
D. 5
E. 6
First of all: \(210 = 2 *3 *5 *7\).
Next:
\(n\) must be a multiple of 2 because we are told that \(\frac{n}{2}\) is an integer;
\(n\) must be a multiple of 3 because we are told that \(\frac{n}{3}\) is an integer;
\(n\) must be a multiple of 5 because either \(\frac{n}{2}\) or \(\frac{n}{3}\) is a multiple of 5 (how else would 5 appear in the LCM?);
\(n\) must be a multiple of 7 because either \(\frac{n}{2}\) or \(\frac{n}{3}\) is a multiple of 7 (how else would 7 appear in the LCM?);
\(n\) cannot have higher powers of 2, 3, 5, or 7 because in this case, the LCM would also contain 2, 3, 5, or 7 in higher power;
\(n\) cannot be a multiple of any other prime because in this case, that prime would also appear in \(\frac{n}{2}\) or \(\frac{n}{3}\) and thus in the LCM.
Thus, \(n\) must be \(210=2*3*5*7\). Therefore, \(n\) has four prime factors.
Answer: C
If the least common multiple of integers \(\frac{n}{2}\) and \(\frac{n}{3}\) is 210, how many distinct prime factors does the positive integer \(n\) have?
A. 2
B. 3
C. 4
D. 5
E. 6
First of all: \(210 = 2 *3 *5 *7\).
Next:
\(n\) must be a multiple of 2 because we are told that \(\frac{n}{2}\) is an integer;
\(n\) must be a multiple of 3 because we are told that \(\frac{n}{3}\) is an integer;
\(n\) must be a multiple of 5 because either \(\frac{n}{2}\) or \(\frac{n}{3}\) is a multiple of 5 (how else would 5 appear in the LCM?);
\(n\) must be a multiple of 7 because either \(\frac{n}{2}\) or \(\frac{n}{3}\) is a multiple of 7 (how else would 7 appear in the LCM?);
\(n\) cannot have higher powers of 2, 3, 5, or 7 because in this case, the LCM would also contain 2, 3, 5, or 7 in higher power;
\(n\) cannot be a multiple of any other prime because in this case, that prime would also appear in \(\frac{n}{2}\) or \(\frac{n}{3}\) and thus in the LCM.
Thus, \(n\) must be \(210=2*3*5*7\). Therefore, \(n\) has four prime factors.
Answer: C
General Discussion
21 May 2024, 02:10
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Bunuel
210 = 7 * 3 * 5 * 2
Hence \(\frac{n}{2}\) and \(\frac{n}{3}\) must have these factors. To make this happen, n should have at 7, 3, 5 and 2 in it.
Distinct factors = 4
Option C
