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31 Jul 2025, 09:28
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If \(x\) is a positive integer and \(x^3\) is a multiple of by 81, which of the following could be the remainder when \(x\) is divided by 36?
I. 0
II. 3
III. 27
A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III
I. 0
II. 3
III. 27
A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III
31 Jul 2025, 09:28
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Official Solution:
If \(x\) is a positive integer and \(x^3\) is a multiple of by 81, which of the following could be the remainder when \(x\) is divided by 36?
I. 0
II. 3
III. 27
A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III
We are told that \(x\) is a positive integer and \(x^3\) is a multiple of \(81 = 3^4\). If \(x^3\) is divisible by \(3^4\), then \(x\) must be divisible by at least \(3^2\). In this case, \(x^3 = (3^2)^3 = 3^6\), which is indeed divisible by \(3^4\). So, the least value of \(x\) is \(3^2 = 9\), and any multiple of 9 would make \(x^3\) divisible by \(3^4\). So, for example, for each of \(3^2\), \(3^2 * 2\), \(3^2 * 3\), ..., \(x^3\) will be divisible by \(3^4\).
Now we're asked which of the following could be the remainder when \(x\) is divided by \(36\). Since \(x\) is a multiple of 9, its remainder when divided by 36 could be:
• 9, in case \(x = 9\)
• 18, in case \(x = 18\)
• 27, in case \(x = 27\)
• 0, in case \(x = 36\)
• 9, again, in case \(x = 45\)
• and so on.
So, from the options, only 0 and 27 could be remainders.
Answer: D
If \(x\) is a positive integer and \(x^3\) is a multiple of by 81, which of the following could be the remainder when \(x\) is divided by 36?
I. 0
II. 3
III. 27
A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III
We are told that \(x\) is a positive integer and \(x^3\) is a multiple of \(81 = 3^4\). If \(x^3\) is divisible by \(3^4\), then \(x\) must be divisible by at least \(3^2\). In this case, \(x^3 = (3^2)^3 = 3^6\), which is indeed divisible by \(3^4\). So, the least value of \(x\) is \(3^2 = 9\), and any multiple of 9 would make \(x^3\) divisible by \(3^4\). So, for example, for each of \(3^2\), \(3^2 * 2\), \(3^2 * 3\), ..., \(x^3\) will be divisible by \(3^4\).
Now we're asked which of the following could be the remainder when \(x\) is divided by \(36\). Since \(x\) is a multiple of 9, its remainder when divided by 36 could be:
• 9, in case \(x = 9\)
• 18, in case \(x = 18\)
• 27, in case \(x = 27\)
• 0, in case \(x = 36\)
• 9, again, in case \(x = 45\)
• and so on.
So, from the options, only 0 and 27 could be remainders.
Answer: D
17 Sep 2025, 08:32
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I did not quite understand the solution. Could you please explain this part of the answer as much of the answer is dependent on it and I can't figure out why this is true. Thanks much.
"If x^3 is divisible by 3^4, then x must be divisible by at least 3^2."
"If x^3 is divisible by 3^4, then x must be divisible by at least 3^2."
