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When m is divided by 7, the remainder is 5. When m is divided by 13
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17 Aug 2016, 11:20
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Re: When m is divided by 7, the remainder is 5. When m is divided by 13
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20 Aug 2016, 09:51
Bunuel wrote: 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 Using Bunuel's formula  m = 7x +5 : 12,19,26 .. m = 13y +6 : 19,... For quotient, LCM (7,13) =91 For remainder, first common value of each pattern i.e. 19 so we can put m as > m = 91q + 19 Now 1 < m < 200 putting q=2 , we see m is 201 so q=1, m =110 Ans E.




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Re: When m is divided by 7, the remainder is 5. When m is divided by 13
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17 Aug 2016, 12:00
Bunuel wrote: 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 Answer is E. I started with E.110 and it fits the bill . 105 is divisible by 7 and 104 is divisible by 13. so it's E.
Modular approach would be the proper way, but this worked at first attempt.



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Re: When m is divided by 7, the remainder is 5. When m is divided by 13
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17 Aug 2016, 12:57
110 is the answer. As 110 is the largest of the five answer choices you can start with it.



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When m is divided by 7, the remainder is 5. When m is divided by 13
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17 Aug 2016, 17:54
first, find the lowest value of m: formula is m=r+dq, or m=remainder+divisor*quotient assume quotient of m/7=2 and quotient of m/13=1 thus, 5+7*2=19 and 6+13*1=19 then, find the next value of m: add the product of two divisors, or 7*13=91, to 19, for a sum of 110 m=110



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Re: When m is divided by 7, the remainder is 5. When m is divided by 13
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17 Aug 2016, 19:44
gracie wrote: first, find the lowest value of m: formula is m=r+dq, or m=remainder+divisor*quotient assume quotient of m/7=2 and quotient of m/13=1 thus, 5+7*2=19 and 6+13*1=19 then, find the next value of m: add the product of two divisors, or 7*13=91, to 19, for a sum of 110 m=110 Thanks gracie, I was trying to recollect this for awhile. It's called Chinese Remainder Theorem.



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When m is divided by 7, the remainder is 5. When m is divided by 13
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17 Aug 2016, 20:26
Senthil1981 wrote: gracie wrote: first, find the lowest value of m: formula is m=r+dq, or m=remainder+divisor*quotient assume quotient of m/7=2 and quotient of m/13=1 thus, 5+7*2=19 and 6+13*1=19 then, find the next value of m: add the product of two divisors, or 7*13=91, to 19, for a sum of 110 m=110 Thanks gracie, I was trying to recollect this for awhile. It's called Chinese Remainder Theorem. Hi Senthil, You're welcome. Here's something I forgot to mention: when you're trying to pick the quotients, remember that the ratio between them will inversely approximate the ratio between divisors. In the problem above, the ratio between divisors is 7:13, while the ratio between quotients is 2:1. I hope this is helpful. gracie



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Re: When m is divided by 7, the remainder is 5. When m is divided by 13
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18 Aug 2016, 10:05
Bunuel wrote: 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 You can also get to the answer by backsolving using the answer choices. Since we want the greatest possible value of m within the given range, let's start with option E Option E 110/7 = 15 + r5 110/13 = 8 + r6 end here, no need to continue! We have gotten what we want. m = 110 Answer: E
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Re: When m is divided by 7, the remainder is 5. When m is divided by 13
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18 Aug 2016, 21:12
Bunuel wrote: 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 Use some logic to solve it orally. Note here that the two divisors are 7 and 13. So when you get the first value of m that satisfies these conditions, you know that you will get all subsequent values by adding 7*13 = 91 to them progressively. So, if 5 were a value of m, there would be other values of m such as 5+91, 5+91+91 etc. Hence, all (A), (B), (C) and (D) cannot be the maximum values of m since when you add 91 to them, you will get a value of m less than 200 and that will be the maximum. Hence, answer has to be (E) only. For more on this concept, check: http://www.veritasprep.com/blog/2011/05 ... spartii/
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Re: When m is divided by 7, the remainder is 5. When m is divided by 13
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05 Sep 2017, 17:03
Bunuel wrote: 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 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 According to the above expression, m can be: 5, 12, 19, ... m = 13z + 6 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
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Re: When m is divided by 7, the remainder is 5. When m is divided by 13
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20 Nov 2018, 17:37
Bunuel wrote: 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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Re: When m is divided by 7, the remainder is 5. When m is divided by 13
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28 Nov 2018, 07:44
Bunuel wrote: 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 We can do backsolving to reach to correct answer here rather than using remainder formula We need to make sure any answer choice which satisfies above 2 condition is the right answer As 110 is largest number in answer choices , start with E 110 when divide by 7 , remainder is 5, 110 when divided by 13, remiander 6, bingo, right answer within 20 seconds
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