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For natural numbers m and n, let k be the remainder when m is divided

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For natural numbers m and n, let k be the remainder when m is divided  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 20 Aug 2019, 12:23
1
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 2019-1000+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  [#permalink]

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New post 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   [#permalink] 22 Aug 2019, 01:00
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