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When m is divided by 7, the remainder is 5. When m is

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When m is divided by 7, the remainder is 5. When m is

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Source: Economist GMAT

When m is divided by 7, the remainder is 5. When m is divided by 13, the remainder is 6. If 1 < m < 200, what is the greatest possible value of m?

A. 5
B. 19
C. 61
D. 74
E. 110

The OA is E.

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Hi All,

We're told that when M is divided by 7, the remainder is 5, when M is divided by 13, the remainder is 6 and 1 < m < 200. We're asked for the GREATEST possible value of M. This question can be solved rather easily by TESTing THE ANSWERS. Let's TEST Answer E first (if it fits what we're told, then we're done; if it does not fit, then we'll Test Answer D and so on).

Answer E: 110
110/7 = 15r5 --> this fits what we're told
110/13 = 8r6 --> this also fits what we're told
Thus, this must be the answer.

Final Answer: E

GMAT assassins aren't born, they're made,
Rich

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Contact Rich at Rich.C@empowergmat.com

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BTGmoderatorLU wrote:
Source: Economist GMAT

When m is divided by 7, the remainder is 5. When m is divided by 13, the remainder is 6. If 1 < m < 200, what is the greatest possible value of m?

A. 5
B. 19
C. 61
D. 74
E. 110
\[? = {m_{\,\max }}\,\,\,\left( {1 < m < 200} \right)\,\,{\text{such}}\,\,{\text{that}}\,\,\,\,\left\{ \begin{gathered}
m = 7Q + 5\,\,\,\left( 1 \right)\,\, \hfill \\
m = 13K + 6\,\,\,\left( 2 \right) \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,\left( {Q,K\,\,\,{\text{ints}}} \right)\]
\[\left\{ \begin{gathered}
\left( 1 \right) \cdot 13\,\,\,\, \Rightarrow \,\,\,\,13m = 7 \cdot 13Q + 65 \hfill \\
\left( 2 \right) \cdot 7\,\,\,\,\,\, \Rightarrow \,\,\,\,7m = 7 \cdot 13K + 42 \hfill \\
\end{gathered} \right.\,\,\,\,\mathop \Rightarrow \limits^{\left( - \right)} \,\,\,\,6m = 7 \cdot 13\left( {Q - K} \right) + 23\]
\[\left\{ \begin{gathered}
\,7m = 7 \cdot 13K + 42 \hfill \\
\,6m = 7 \cdot 13\left( {Q - K} \right) + 23 \hfill \\
\end{gathered} \right.\,\,\,\,\,\mathop \Rightarrow \limits^{\left( - \right)} \,\,\,\,m = 7 \cdot 13\left( {2K - Q} \right) + 19\]
\[\left\{ \begin{gathered}
\,m = 91J + 19 \hfill \\
\,1 < m < 200 \hfill \\
\end{gathered} \right.\,\,\,\,\left( {J\,\,\operatorname{int} } \right)\,\,\,\, \Rightarrow \,\,\,\,\,{J_{\max }} = 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = {m_{\,\max }} = 91 + 19 = 110\]

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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Portuguese-speakers :: https://www.gmath.com.br

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BTGmoderatorLU wrote:
Source: Economist GMAT

When m is divided by 7, the remainder is 5. When m is divided by 13, the remainder is 6. If 1 < m < 200, what is the greatest possible value of m?

A. 5
B. 19
C. 61
D. 74
E. 110

The OA is E.
We are given two properties about m: when m is divided by 7, the remainder is 5; further, when m is divided by 13, the remainder is 6. We can create the following two equations:

m = 7q+ 5 for some nonnegative integer q

According to the above expression, m can be:

5, 12, 19, ...

m = 13z + 6 for some nonnegative integer z

According to the above expression, m can be:

6, 19, 32, …

We can see that 19 is the smallest positive integer value of m that satisfies the properties. However, we are asked to find the largest integer less than 200 that satisfies the properties. In that case, we can add 19 to any number that is both divisible by 13 and 7, in other words, a number that is a multiple of both 13 and 7. Since the LCM of 13 and 7 is 13 x 7 = 91, the next value of m is 19 + 91 = 110, which happens to be the largest possible value of m that is still less than 200. (Note: the next value of m is 110 + 91 = 201, which is greater than 200.)

Alternate solution:

The problem is asking for the largest possible value less than 200 that satisfies the following properties: when m is divided by 7, the remainder is 5, and when m is divided by 13, the remainder is 6. We can check the largest number in the given answer choices first and work backward until we find the answer. So let’s check 110 first:

110/7 = 15 R 5 and 110/13 = 8 R 6

We see that 110 satisfies both properties, so 110 is the answer.
.
Answer: E

_________________

Scott Woodbury-Stewart
Founder and CEO
scott@targettestprep.com



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