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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7875
GMAT 1: 760 Q51 V42 GPA: 3.82
For natural numbers m and n, let k be the remainder when m is divided  [#permalink]

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Difficulty:   85% (hard)

Question Stats: 46% (03:29) correct 54% (02:57) wrong based on 24 sessions

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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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Director  D
Joined: 19 Oct 2018
Posts: 852
Location: India
Re: For natural numbers m and n, let k be the remainder when m is divided  [#permalink]

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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

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 $$f(2^{1000}, 7)+ f(2^{1001}, 7)+ f(2^{1002}, 7)+…+ f(2^{2019}, 7)$$?

$$A. \frac{1019}{7}$$

$$B. \frac{1020}{7}$$

$$C. 240$$

$$D. 340$$

$$E. 441$$
SVP  P
Joined: 03 Jun 2019
Posts: 1500
Location: India
For natural numbers m and n, let k be the remainder when m is divided  [#permalink]

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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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Joined: 03 Jun 2019
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For natural numbers m and n, let k be the remainder when m is divided  [#permalink]

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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
_________________
"Success is not final; failure is not fatal: It is the courage to continue that counts."

Please provide kudos if you like my post. Kudos encourage active discussions.

My GMAT Resources: -

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All you need to know about GMAT quant

Tele: +91-11-40396815
Mobile : +91-9910661622
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Manager  B
Joined: 20 Jul 2012
Posts: 118
GMAT 1: 650 Q47 V33 Re: For natural numbers m and n, let k be the remainder when m is divided  [#permalink]

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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.
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7875
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: For natural numbers m and n, let k be the remainder when m is divided  [#permalink]

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=>

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.
_________________ 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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