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16 Sep 2014, 01:00
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If \(m\) and \(n\) are positive integers, is the remainder of \(\frac{10^m + n}{3}\) greater than the remainder of \(\frac{10^n + m}{3}\)?
(1) \(m \gt n\)
(2) The remainder of \(\frac{n}{3}\) is \(2\)
(1) \(m \gt n\)
(2) The remainder of \(\frac{n}{3}\) is \(2\)
16 Sep 2014, 01:00
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Official Solution:
If \(m\) and \(n\) are positive integers, is the remainder of \(\frac{10^m + n}{3}\) greater than the remainder of \(\frac{10^n + m}{3}\)?
Note: This is a difficult question that requires careful reading and understanding of the solution. It's important to take your time and fully comprehend the reasoning behind each step in the solution. Don't hesitate to ask for clarification if needed.
Firstly, it is important to note that any positive integer can only have three possible remainders upon division by 3: 0, 1, or 2.
Given that the sum of the digits of \(10^m\) and \(10^n\) is always equal to 1, the remainders of \(\frac{10^m + n}{3}\) and \(\frac{10^n + m}{3}\) depend only on the value of the number added to \(10^m\) and \(10^n\). There are three possible cases:
(1) \(m \gt n\). Not sufficient.
(2) The remainder of \(\frac{n}{3}\) is \(2\).
The above implies that \(n\) can take any of the values 2, 5, 8, 11, and so on. When any such value of \(n\) is added to \(10^m\), the resulting number has digits that add up to a multiple of 3, which means that its remainder when divided by 3 is 0. Thus, the remainder of \(\frac{10^m + n}{3}\) is 0.
Now, the question asks whether the remainder of \(\frac{10^m + n}{3}\), which is 0, is greater than the remainder of \(\frac{10^n + m}{3}\), which can only be 0, 1, or 2. It is clear that the remainder of \(\frac{10^m + n}{3}\) cannot be greater than the remainder of \(\frac{10^n + m}{3}\), it may only be less than or equal to it. Therefore, the answer to the question is NO. This statement alone is sufficient.
Answer: B
If \(m\) and \(n\) are positive integers, is the remainder of \(\frac{10^m + n}{3}\) greater than the remainder of \(\frac{10^n + m}{3}\)?
Note: This is a difficult question that requires careful reading and understanding of the solution. It's important to take your time and fully comprehend the reasoning behind each step in the solution. Don't hesitate to ask for clarification if needed.
Firstly, it is important to note that any positive integer can only have three possible remainders upon division by 3: 0, 1, or 2.
Given that the sum of the digits of \(10^m\) and \(10^n\) is always equal to 1, the remainders of \(\frac{10^m + n}{3}\) and \(\frac{10^n + m}{3}\) depend only on the value of the number added to \(10^m\) and \(10^n\). There are three possible cases:
- • If the number added to \(10^m\) and \(10^n\) is a multiple of 3 (i.e., 0, 3, 6, 9, ...), then the remainder when \(10^m+n\) or \(10^n+m\) is divided by 3 will be 1. This is because the sum of the digits of \(10^m+n\) and \(10^n+m\) will be 1 more than a multiple of 3.
• If the number added to \(10^m\) and \(10^n\) is one more than a multiple of 3 (i.e., 1, 4, 7, 10, ...), then the remainder when \(10^m+n\) or \(10^n+m\) is divided by 3 will be 2. This is because the sum of the digits of \(10^m+n\) and \(10^n+m\) will be 2 more than a multiple of 3.
• If the number added to \(10^m\) and \(10^n\) is two more than a multiple of 3 (i.e., 2, 5, 8, 11, ...), then the remainder when \(10^m+n\) or \(10^n+m\) is divided by 3 will be 0. This is because the sum of the digits of \(10^m+n\) and \(10^n+m\) will be a multiple of 3.
(1) \(m \gt n\). Not sufficient.
(2) The remainder of \(\frac{n}{3}\) is \(2\).
The above implies that \(n\) can take any of the values 2, 5, 8, 11, and so on. When any such value of \(n\) is added to \(10^m\), the resulting number has digits that add up to a multiple of 3, which means that its remainder when divided by 3 is 0. Thus, the remainder of \(\frac{10^m + n}{3}\) is 0.
Now, the question asks whether the remainder of \(\frac{10^m + n}{3}\), which is 0, is greater than the remainder of \(\frac{10^n + m}{3}\), which can only be 0, 1, or 2. It is clear that the remainder of \(\frac{10^m + n}{3}\) cannot be greater than the remainder of \(\frac{10^n + m}{3}\), it may only be less than or equal to it. Therefore, the answer to the question is NO. This statement alone is sufficient.
Answer: B
04 Jul 2017, 12:33
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One could alternatively solve this question by testing some values
So WKT m,n are positive integers. Question - Is rem of \(\frac{(10^m + n)}{3}\) > rem of \(\frac{(10^n + m)}{3}\)? Yes/No
1) m>n
Say m = 2 and n = 1
rem of \(\frac{(10^2 + 1)}{3}\) = 2
rem of \(\frac{(10^1 + 2)}{3}\) = 0
YES
[Trying to disprove]
Say m = 3 and n = 2
rem of \(\frac{(10^3 + 2)}{3}\) = 0
rem of \(\frac{(10^2 + 3)}{3}\) = 1
NO
Hence Insuff
2) This is interesting n/3 = __r2
n = 2, 5, 8, 11....
Notice how this effects the rem of \(\frac{(10^m + n)}{3}\). No matter what you plug in for m the rem is ALWAYS 0. And since the rem CANNOT be neg numbers, rem of \(\frac{(10^n + m)}{3}\) can be 0 or greater, leading to ALWAYS NO answer.
Hence suff.
Option B
So WKT m,n are positive integers. Question - Is rem of \(\frac{(10^m + n)}{3}\) > rem of \(\frac{(10^n + m)}{3}\)? Yes/No
1) m>n
Say m = 2 and n = 1
rem of \(\frac{(10^2 + 1)}{3}\) = 2
rem of \(\frac{(10^1 + 2)}{3}\) = 0
YES
[Trying to disprove]
Say m = 3 and n = 2
rem of \(\frac{(10^3 + 2)}{3}\) = 0
rem of \(\frac{(10^2 + 3)}{3}\) = 1
NO
Hence Insuff
2) This is interesting n/3 = __r2
n = 2, 5, 8, 11....
Notice how this effects the rem of \(\frac{(10^m + n)}{3}\). No matter what you plug in for m the rem is ALWAYS 0. And since the rem CANNOT be neg numbers, rem of \(\frac{(10^n + m)}{3}\) can be 0 or greater, leading to ALWAYS NO answer.
Hence suff.
Option B
