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When N is divisible by 2 and 8 means, N has at least 3 2s in it (because 8 = 2x2x2). With this information given, look at the options: 1. Is N divisible by 4 --- Yes, because 4 = 2x2 2. Is N divisible by 6 --- Don't know for sure, because 6 = 2x3. From the given information, we don't know if N has a 3 in it. 3. Is N divisible by 8 --- Don't know again, because 8 = 2x2x2x2. From the given information again, we don't know if N has 4 2s in it.
Because this is a "Must" question, we know for sure that N is divisible by 4.
Am not sure If I got the correctly. If we assume N=48. Then It would be divisible by 4, 6 and 16. Then options dont make sense. Please help me understand the question. where am going wrong?
Am not sure If I got the correctly. If we assume N=48. Then It would be divisible by 4, 6 and 16. Then options dont make sense. Please help me understand the question. where am going wrong?
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If a number N is divisible by both 2 and 8, then which of the following statements must be true?
I. N is divisible by 4 II. N is divisible by 6 III. N is divisible by 16
A. I only B. II only C. III only D. I and II only E. I and III only
Notice that the question asks "which of the following statements must be true", not "which of the following statements could be true". While N could be divisible by any number, it must be divisible only by some of the numbers.
Generally if a positive integer x is divisible by positive integers a and b, then it will be divisible by the least common multiple of a and b. Thus since N is divisible by both 2 and 8, then it will be divisible by the LCM(2, 8), which is 8. If N is divisible by 8, then it's divisible by every factor of 8: 1, 2, 4, and 8. Therefore, only option A MUST be true.
Am not sure If I got the correctly. If we assume N=48. Then It would be divisible by 4, 6 and 16. Then options dont make sense. Please help me understand the question. where am going wrong?
If a number N is divisible by both 2 and 8, then which of the following statements must be true?
I. N is divisible by 4 II. N is divisible by 6 III. N is divisible by 16
A. I only B. II only C. III only D. I and II only E. I and III only
Notice that the question asks "which of the following statements must be true", not "which of the following statements could be true". While N could be divisible by any number, it must be divisible only by some of the numbers.
Generally if a positive integer x is divisible by positive integers a and b, then it will be divisible by the least common multiple of a and b. Thus since N is divisible by both 2 and 8, then it will be divisible by the LCM(2, 8), which is 8. If N is divisible by 8, then it's divisible by every factor of 8: 1, 2, 4, and 8. Therefore, only option A MUST be true.
Am not sure If I got the correctly. If we assume N=48. Then It would be divisible by 4, 6 and 16. Then options dont make sense. Please help me understand the question. where am going wrong?
If a number N is divisible by both 2 and 8, then which of the following statements must be true?
I. N is divisible by 4 II. N is divisible by 6 III. N is divisible by 16
A. I only B. II only C. III only D. I and II only E. I and III only
Notice that the question asks "which of the following statements must be true", not "which of the following statements could be true". While N could be divisible by any number, it must be divisible only by some of the numbers.
Generally if a positive integer x is divisible by positive integers a and b, then it will be divisible by the least common multiple of a and b. Thus since N is divisible by both 2 and 8, then it will be divisible by the LCM(2, 8), which is 8. If N is divisible by 8, then it's divisible by every factor of 8: 1, 2, 4, and 8. Therefore, only option A MUST be true.
Answer: A.
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Adding on to what Bunuel explained, we can write \(N = 8K\)
Now, \(\frac{N}{4} = \frac{8k}{4} =\) Always an Integer
\(\frac{N}{6} = \frac{8k}{6} =\) May or may not be an integer, depending upon the value of k (if \(k=1\), \(\frac{N}{6}\) is not an integer, but if \(k=3\), \(\frac{N}{6}\) is an integer)
\(\frac{N}{16} = \frac{8k}{16} =\) May or may not be an integer, depending upon the value of k (if \(k=1\), \(\frac{N}{16}\) is not an integer, but if \(k=2\), \(\frac{N}{16}\) is an integer)
Hence only I must be true . Option A
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