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

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

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20 Aug 2019, 00:49
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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 $$f(2^{1000}, 7)+ f(2^{1001}, 7)+ f(2^{1002}, 7)+…+ f(2^{2019}, 7)$$?

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

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

$$C. 240$$

$$D. 340$$

$$E. 441$$

_________________
MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Director 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] ### Show Tags 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 $$f(2^{1000}, 7)+ f(2^{1001}, 7)+ f(2^{1002}, 7)+…+ f(2^{2019}, 7)$$? $$A. \frac{1019}{7}$$ $$B. \frac{1020}{7}$$ $$C. 240$$ $$D. 340$$ $$E. 441$$ SVP Joined: 03 Jun 2019 Posts: 1500 Location: India For natural numbers m and n, let k be the remainder when m is divided [#permalink] ### Show Tags 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 _________________ "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: - Efficient Learning All you need to know about GMAT quant Tele: +91-11-40396815 Mobile : +91-9910661622 E-mail : kinshook.chaturvedi@gmail.com 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. SVP Joined: 03 Jun 2019 Posts: 1500 Location: India For natural numbers m and n, let k be the remainder when m is divided [#permalink] ### Show Tags 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 _________________ "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: - Efficient Learning All you need to know about GMAT quant Tele: +91-11-40396815 Mobile : +91-9910661622 E-mail : kinshook.chaturvedi@gmail.com Manager 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] ### Show Tags 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. Math Revolution GMAT Instructor 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] ### Show Tags 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 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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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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