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Bunuel
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Bunuel
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I think this question is good and helpful.
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I think this is a high-quality question and I agree with explanation. Given, m,n are positive numbers. So taking '0' in to calculation is wrong, I guess. In the step... value of m,n be 0,3,6,9,.................
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I think this is a high-quality question and I agree with explanation. Quite helpful.
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I think this is a high-quality question and I agree with explanation.
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Bunuel
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 \(2\)

Target question: Is the remainder of \(\frac{10^m + n}{3}\) greater than the remainder of \(\frac{10^n + m}{3}\)?

Statement 1: \(m \gt n\)
This statement doesn't feel sufficient, so I'll TEST some values.
There are several values of m and n that satisfy statement 1. Here are two:
Case a: m = 2 and n = 1. \(10^m + n=10^2 + 1=101\), and 101 divided by 3 leaves a remainder of 2. Similarly, \(10^n + m=10^1 + 2=12\), and 12 divided by 3 leaves a remainder of 0. So, in this case, the answer to the target question is YES
Case b: m = 3 and n = 2. \(10^m + n=10^3 + 2=1002\), and 1002 divided by 3 leaves a remainder of 0. Similarly, \(10^n + m=10^2 + 3=103\), and 103 divided by 3 leaves a remainder of 1. So, in this case, the answer to the target question is NO
Since we can’t answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: The remainder of \(\frac{n}{3}\) is \(2\)
In other words, n is 2 greater than some of multiple of 3
In other words, n = 3k + 2 for some integer k

We must also recognize that when we divide \(10^m - 1\) by 3, the remainder will always be 0 (as long as b is a positive integer)
We know this because \(10^m - 1\) will always result in an integer that consists solely of 9's.
For example, \(10^2 - 1=99\) and \(10^4 - 1=9999\) etc.
The divisibility rule for 3 tells us that any number consisting solely of 9's will be divisible by 3.
If \(10^m - 1\) is divisible by 3, then \(10^m\) is 1 greater than a multiple of 3
In other words \(10^m\) = 3j + 1 for some integer j.

At this point, we can see that \(10^m + n = (3j + 1) + (3k + 2)=3j+3k+3=3(j+k+1)\)
Since we can be certain that \(3(j+k+1)\) is a multiple of 3, we know that 3(j+k+1) divided by 3 must leave a remainder of 0
Since 0 is the smallest possible remainder, we can be certain that the remainder of \(\frac{10^m + n}{3}\) is NOT greater than the remainder of \(\frac{10^n + m}{3}\)

Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: B

Cheers,
Brent
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I think this is a high-quality question and I agree with explanation.
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I think this is a high-quality question and I agree with explanation.
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In B how can we esfimate ghe value of M ??? If M is equal to N or less than N or Negative terms than??

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Sarmadk5
In B how can we esfimate ghe value of M ??? If M is equal to N or less than N or Negative terms than??

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The question asks: is the remainder of \(\frac{10^m + n}{3}\) greater than the remainder of \(\frac{10^n + m}{3}\)?

From (2) we get that the remainder of \(\frac{10^m + n}{3}\) is 0, which is the minimum possible remainder, so it cannot be greater than the reminder of \(\frac{10^n + m}{3}\), whatever it might be.
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Bunuel I'm trying to understand your answer while applying it to other use cases. For example you said:

Quote:
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

This rule works for dividing by 3. However it doesnt seem to work for dividing by 4. For example:
- 4 has 4 possible remaineders: 0, 1, 2, 3
- 32 / 4 --> sum of digits is 5, which is +1 of a multiple of 4. but the remainder is 0
- 33 / 4 --> sum of digits is 6, which is +2 of a multiple of 4, but the remainder is 1
- etc

Is there something i'm missing about trying to generalize this rule? how did we land on this method of thinking specifically for the 3 use case?
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Bunuel
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:

• 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


Bunuel I'm trying to understand your answer while applying it to other use cases. For example you said:

Quote:
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

This rule works for dividing by 3. However it doesnt seem to work for dividing by 4. For example:
- 4 has 4 possible remaineders: 0, 1, 2, 3
- 32 / 4 --> sum of digits is 5, which is +1 of a multiple of 4. but the remainder is 0
- 33 / 4 --> sum of digits is 6, which is +2 of a multiple of 4, but the remainder is 1
- etc

Is there something i'm missing about trying to generalize this rule? how did we land on this method of thinking specifically for the 3 use case?

Not entirely sure what you are trying to generalize there, but the divisibility rule for 3 (and 9) is distinct from divisibility rules for other numbers. A number is divisible by 3 if the sum of its digits is divisible by 3. When the sum of a number's digits is 1 more than a multiple of 3, dividing that number by 3 results in a remainder of 1. If the sum of a number's digits is 2 more than a multiple of 3, dividing that number by 3 results in a remainder of 2. This principle applies to 3 and 9 but not for numbers like 4.
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Bunuel yes, that answers my question. I wasnt aware that this rule only applies to 3's / 9's. I was trying to prove if it is a general rule of division / remainders and it seemed like it wasnt, so just wanted to confirm
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