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Math Revolution GMAT Instructor
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For natural numbers m and n, let k be the remainder when m is divided
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20 Aug 2019, 00:49
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[GMAT math practice question] For natural numbers \(m\) and \(n\), let \(k\) be the remainder when \(m\) is divided by \(n\). In that case, we define \(f(m, n)\) as \(\frac{k}{n}.\) What is the value of \(f(2^{1000}, 7)+ f(2^{1001}, 7)+ f(2^{1002}, 7)+…+ f(2^{2019}, 7)\)? \(A. \frac{1019}{7}\) \(B. \frac{1020}{7}\) \(C. 240\) \(D. 340\) \(E. 441\)
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Re: For natural numbers m and n, let k be the remainder when m is divided
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20 Aug 2019, 05:13
\(2^3\)= 1 mod 7 \((2^3)^{333}\)=\(2^{999}\)=1 mod 7 \(2^{1000}\)= 2 mod 7 \(2^{1001}\)=(2*2) mod 7= 4 mod 7 \(2^{1002}\)= (2*4) mod 7= 1 mod 7 ..... so on \(f(2^{1000}, 7)+ f(2^{1001}, 7)+ f(2^{1002}, 7)+…+ f(2^{2019}, 7)\) = {2+4+1+2+4+1.....1020 terms....2+4+1}/7 = 1*1020/3 =340 MathRevolution wrote: [GMAT math practice question]
For natural numbers \(m\) and \(n\), let \(k\) be the remainder when \(m\) is divided by \(n\). In that case, we define \(f(m, n)\) as \(\frac{k}{n}.\) What is the value of \(f(2^{1000}, 7)+ f(2^{1001}, 7)+ f(2^{1002}, 7)+…+ f(2^{2019}, 7)\)?
\(A. \frac{1019}{7}\) \(B. \frac{1020}{7}\) \(C. 240\) \(D. 340\) \(E. 441\)



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For natural numbers m and n, let k be the remainder when m is divided
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Updated on: 20 Aug 2019, 06:25
MathRevolution wrote: [GMAT math practice question]
For natural numbers \(m\) and \(n\), let \(k\) be the remainder when \(m\) is divided by \(n\). In that case, we define \(f(m, n)\) as \(\frac{k}{n}.\) What is the value of \(f(2^{1000}, 7)+ f(2^{1001}, 7)+ f(2^{1002}, 7)+…+ f(2^{2019}, 7)\)?
\(A. \frac{1019}{7}\) \(B. \frac{1020}{7}\) \(C. 240\) \(D. 340\) \(E. 441\) Given: For natural numbers \(m\) and \(n\), let \(k\) be the remainder when \(m\) is divided by \(n\). In that case, we define \(f(m, n)\) as \(\frac{k}{n}.\) Asked: What is the value of \(f(2^{1000}, 7)+ f(2^{1001}, 7)+ f(2^{1002}, 7)+…+ f(2^{2019}, 7)\)? \(f(2^{1000}, 7) = Rem [(8)^{333} *2,7] /7 = 2/7\) \(f(2^{1001}, 7) = Rem [(8)^{333} *2^2,7] /7 = 4/7\) \(f(2^{1002}, 7) = Rem [(8)^{334} ,7] /7 = 1/7\) \(f(2^{1003}, 7) = Rem [(8)^{334}*2 ,7] /7 = 2/7\) \(f(2^{1004}, 7) = Rem [(8)^{334}*2^2 ,7] /7 = 4/7\) No of such terms = 2019  1000 + 1= 1020 = 3*340 S = \(f(2^{1000}, 7)+ f(2^{1001}, 7)+ f(2^{1002}, 7)+…+ f(2^{2019}, 7)\) \(=\frac{1}{7} * (2 + 4 + 1 + 2 + 4 + 1 ..... 1020 terms)\) \(= \frac{1}{7} * 7*340 = 340\) IMO D
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Originally posted by Kinshook on 20 Aug 2019, 06:16.
Last edited by Kinshook on 20 Aug 2019, 06:25, edited 2 times in total.



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For natural numbers m and n, let k be the remainder when m is divided
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20 Aug 2019, 06:19
Kinshook wrote: MathRevolution wrote: [GMAT math practice question]
For natural numbers \(m\) and \(n\), let \(k\) be the remainder when \(m\) is divided by \(n\). In that case, we define \(f(m, n)\) as \(\frac{k}{n}.\) What is the value of \(f(2^{1000}, 7)+ f(2^{1001}, 7)+ f(2^{1002}, 7)+…+ f(2^{2019}, 7)\)?
\(A. \frac{1019}{7}\) \(B. \frac{1020}{7}\) \(C. 240\) \(D. 340\) \(E. 441\) Given: For natural numbers \(m\) and \(n\), let \(k\) be the remainder when \(m\) is divided by \(n\). In that case, we define \(f(m, n)\) as \(\frac{k}{n}.\) Asked: What is the value of \(f(2^{1000}, 7)+ f(2^{1001}, 7)+ f(2^{1002}, 7)+…+ f(2^{2019}, 7)\)? \(f(2^{1000}, 7) = Rem \{(8)^{333} *2,7\} /7 = 2/7\) \(f(2^{1001}, 7) = Rem \{(8)^{333} *2^2,7\} /7 = 4/7\) \(f(2^{1002}, 7) = Rem \{(8)^{334} ,7\} /7 = 1/7\) \(f(2^{1003}, 7) = Rem \{(8)^{334}*2 ,7\} /7 = 2/7\) \(f(2^{1004}, 7) = Rem \{(8)^{334}*2^2 ,7\} /7 = 4/7\) No of such terms = 2019  1000 + 1= 1020 = 3*340 S = \(f(2^{1000}, 7)+ f(2^{1001}, 7)+ f(2^{1002}, 7)+…+ f(2^{2019}, 7)\) \(=\frac{1}{7} * (2 + 4 + 1 + 2 + 4 + 1 ..... 1020 terms)\) \(= \frac{1}{7} * 7*340 = 340\) Since 2+4+1=7 & 1020=3*340 IMO D
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Re: For natural numbers m and n, let k be the remainder when m is divided
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20 Aug 2019, 12:23
MathRevolution wrote: [GMAT math practice question]
For natural numbers \(m\) and \(n\), let \(k\) be the remainder when \(m\) is divided by \(n\). In that case, we define \(f(m, n)\) as \(\frac{k}{n}.\) What is the value of \(f(2^{1000}, 7)+ f(2^{1001}, 7)+ f(2^{1002}, 7)+…+ f(2^{2019}, 7)\)?
\(A. \frac{1019}{7}\) \(B. \frac{1020}{7}\) \(C. 240\) \(D. 340\) \(E. 441\) 2^1/7 gives remainder as 2 2^2/7 gives remainder 4 2^3/7 gives remainder 1 2^4/7 gives remainder 2.... So the remainder follows the series of 2,4,1,2,4,1,2,4,1....... For 2^1000/7, remainder will be 2. fn = 2/7 For 2^1001/7, remainder will be 4 fn = 4/7 For 2^1002/7, remainder will be 1. fn = 1/7 Adding these 3 will give 2/7+4/7+1/7 = 1. So every three terms will add to 1. From 1000 till 2019 there will be 20191000+1 terms = 1020 terms. In 1020 terms there will be 1020/3 set of three terms which add upto 1. = 340 terms = 340*1 = 340. Hence D.



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Re: For natural numbers m and n, let k be the remainder when m is divided
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22 Aug 2019, 01:00
=> Since \(2^1 = 7*0 + 2, 2^2 = 7*0 + 4, 2^3 = 7*1 + 1, 2^4 = 16 = 7*2 + 2, 2^5 = 32 = 7*4 + 4, 2^6 = 64 = 7*9 + 1, …\), the remainders form a pattern which repeats \(2, 4\) and \(1.\) Then, \(f(2^{1000}, 7) = \frac{2}{7}, f(2^{1001}, 7) = \frac{4}{7}, f(2^{1002}, 7) = \frac{1}{7}, …,\) and \(f(2^{2019}, 7) = \frac{1}{7}.\) There are \(1020\) terms \(f(2^{1000}, 7), f(2^{1001}, 7), f(2^{1002}, 7), … , f(2^{2019}, 7).\) So, \(f(2^{1000}, 7)+ f(2^{1001}, 7)+ f(2^{1002}, 7)+…+ f(2^{2019}, 7) = (\frac{2}{7} + \frac{4}{7} + \frac{1}{7})*(\frac{1020}{3}) = 340.\) Therefore, D is the answer. Answer: D
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Re: For natural numbers m and n, let k be the remainder when m is divided
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