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$M$ and $N$, are positive integers that have remainders of 3 and 2 res
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27 Jun 2025, 05:54
27 Jun 2025, 05:54
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$\(M\)$ and $\(N\)$, are positive integers that have remainders of 3 and 2 respectively, when divided by 6 . Which of the following could be the remainder when $\((M+N)\)$ is divided by 12 ?
\(\boxed{A} - 3\)
\(\boxed{B} - 5\)
\(\boxed{C} - 11\)
\(\boxed{A} - 3\)
\(\boxed{B} - 5\)
\(\boxed{C} - 11\)
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$M$ and $N$, are positive integers that have remainders of 3 and 2 res
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28 Jun 2025, 01:02
28 Jun 2025, 01:02
1
Let a be the quotient for M when divided by 6 and b the quotient for N
M = 6a + 3
N = 6b + 2
M + N = 6( a + b ) + 5
Since a + b can only be an integer let the value = c
Therefore M + N = 6c + 5
If c = 0
M + N = 5, which gives a remainder of 5 when divided by 12
If c = 1
M + N = 11, which leaves a remainder of 11 when divided by 12
If c = 2
M + N = 17, which gives 5 when divided by 12
If c = 3
M + N = 23, gives 11 when divided by 12
So we can clearly see a pattern of remainder that is 5 and 11
Answer = B and C
M = 6a + 3
N = 6b + 2
M + N = 6( a + b ) + 5
Since a + b can only be an integer let the value = c
Therefore M + N = 6c + 5
If c = 0
M + N = 5, which gives a remainder of 5 when divided by 12
If c = 1
M + N = 11, which leaves a remainder of 11 when divided by 12
If c = 2
M + N = 17, which gives 5 when divided by 12
If c = 3
M + N = 23, gives 11 when divided by 12
So we can clearly see a pattern of remainder that is 5 and 11
Answer = B and C
Re: $M$ and $N$, are positive integers that have remainders of 3 and 2 res
[#permalink]
28 Jun 2025, 04:00
28 Jun 2025, 04:00
Expert Reply
To solve this problem, we need to analyze the properties of $M$ and $N$ based on their remainders when divided by 6 , and then find the possible remainders of their sum when divided by 12 .
1. Express $M$ and $N$ in terms of the given remainders:
- For $M$ : When $M$ is divided by 6 , the remainder is 3 .
We can write $\(M=6 k+3\)$, where $k$ is a non-negative integer.
(Examples of $\(\mathrm{M}: 3,9,15,21, \ldots\)$ )
- For $N$ : When $N$ is divided by 6 , the remainder is 2 .
We can write $\(N=6 j+2\)$, where $j$ is a non-negative integer.
(Examples of $\(\mathrm{N}: 2,8,14,20, \ldots\)$ )
2. Find the expression for $(M+N)$ :
Add the expressions for $M$ and $N$ :
$$
\(\begin{aligned}
& M+N=(6 k+3)+(6 j+2) \\
& M+N=6 k+6 j+5 \\
& M+N=6(k+j)+5
\end{aligned}\)
$$
Let $X=k+j$. Since $k$ and $j$ are non-negative integers, $X$ can be any non-negative integer ( $\(0,1,2,3, \ldots\))$.
So, $\(M+N=6 X+5\)$.
3. Determine the remainder when $(M+N)$ is divided by $\(\mathbf{1 2}\)$ :
We need to find $(6 X+5)(\bmod 12)$.
The value of $X$ will determine the remainder. We need to consider two cases for $X$ :
- Case 1: $X$ is an even integer.
Let $X=2 Y$ for some non-negative integer $Y$.
Substitute $X$ into the expression for $M+N$ :
$$
\(\begin{aligned}
& M+N=6(2 Y)+5 \\
& M+N=12 Y+5
\end{aligned}\)
$$
When $12 Y+5$ is divided by 12 , the remainder is 5 .
- Case 2: $X$ is an odd integer.
Let $X=2 Y+1$ for some non-negative integer $Y$.
Substitute $X$ into the expression for $M+N$ :
$$
\(\begin{aligned}
& M+N=6(2 Y+1)+5 \\
& M+N=12 Y+6+5 \\
& M+N=12 Y+11
\end{aligned}\)
$$
When $12 Y+11$ is divided by 12 , the remainder is 11 .
Conclusion:
The possible remainders when $(M+N)$ is divided by 12 are 5 and 11 .
Now, let's check the given options:
(A) 3
(B) 5
(C) 11
Both 5 and 11 are possible remainders. Since this is a "multiple answer choices" question (usually indicated by square checkboxes in the actual GRE exam, though shown as (A), (B), (C) here), you should select all options that could be true.
The final answer is $\(\mathrm{B}, \mathrm{C}\)$.
1. Express $M$ and $N$ in terms of the given remainders:
- For $M$ : When $M$ is divided by 6 , the remainder is 3 .
We can write $\(M=6 k+3\)$, where $k$ is a non-negative integer.
(Examples of $\(\mathrm{M}: 3,9,15,21, \ldots\)$ )
- For $N$ : When $N$ is divided by 6 , the remainder is 2 .
We can write $\(N=6 j+2\)$, where $j$ is a non-negative integer.
(Examples of $\(\mathrm{N}: 2,8,14,20, \ldots\)$ )
2. Find the expression for $(M+N)$ :
Add the expressions for $M$ and $N$ :
$$
\(\begin{aligned}
& M+N=(6 k+3)+(6 j+2) \\
& M+N=6 k+6 j+5 \\
& M+N=6(k+j)+5
\end{aligned}\)
$$
Let $X=k+j$. Since $k$ and $j$ are non-negative integers, $X$ can be any non-negative integer ( $\(0,1,2,3, \ldots\))$.
So, $\(M+N=6 X+5\)$.
3. Determine the remainder when $(M+N)$ is divided by $\(\mathbf{1 2}\)$ :
We need to find $(6 X+5)(\bmod 12)$.
The value of $X$ will determine the remainder. We need to consider two cases for $X$ :
- Case 1: $X$ is an even integer.
Let $X=2 Y$ for some non-negative integer $Y$.
Substitute $X$ into the expression for $M+N$ :
$$
\(\begin{aligned}
& M+N=6(2 Y)+5 \\
& M+N=12 Y+5
\end{aligned}\)
$$
When $12 Y+5$ is divided by 12 , the remainder is 5 .
- Case 2: $X$ is an odd integer.
Let $X=2 Y+1$ for some non-negative integer $Y$.
Substitute $X$ into the expression for $M+N$ :
$$
\(\begin{aligned}
& M+N=6(2 Y+1)+5 \\
& M+N=12 Y+6+5 \\
& M+N=12 Y+11
\end{aligned}\)
$$
When $12 Y+11$ is divided by 12 , the remainder is 11 .
Conclusion:
The possible remainders when $(M+N)$ is divided by 12 are 5 and 11 .
Now, let's check the given options:
(A) 3
(B) 5
(C) 11
Both 5 and 11 are possible remainders. Since this is a "multiple answer choices" question (usually indicated by square checkboxes in the actual GRE exam, though shown as (A), (B), (C) here), you should select all options that could be true.
The final answer is $\(\mathrm{B}, \mathrm{C}\)$.
