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

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

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.

=>

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