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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 \(f(2^{1000}, 7)+ f(2^{1001}, 7)+ f(2^{1002}, 7)+…+ f(2^{2019}, 7)\)?
\(A. \frac{1019}{7}\)
\(B. \frac{1020}{7}\)
\(C. 240\)
\(D. 340\)
\(E. 441\)
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 \(f(2^{1000}, 7)+ f(2^{1001}, 7)+ f(2^{1002}, 7)+…+ f(2^{2019}, 7)\)?
\(A. \frac{1019}{7}\)
\(B. \frac{1020}{7}\)
\(C. 240\)
\(D. 340\)
\(E. 441\)
20 Aug 2019, 05:13
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\(2^3\)= 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
\((2^3)^{333}\)=\(2^{999}\)=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
MathRevolution
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
