BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

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

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Elite Legendary Member
Posts: 3991
Joined: Fri Jul 24, 2015 2:28 am
Location: Las Vegas, USA
Thanked: 19 times
Followed by:37 members

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

by Max@Math Revolution » Mon Aug 19, 2019 11:32 pm

Timer

00:00

Your Answer

A

B

C

D

E

Global Stats

[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 k/n.
What is the value of f(2^{1000}, 7)+ f(2^{1001}, 7)+ f(2^{1002}, 7)+...+ f(2^{2019}, 7)?

A. 1019/7
B. 1020/7
C. 240
D. 340
E. 441
Top
Source: — Problem Solving |
User avatar
Elite Legendary Member
Posts: 3991
Joined: Fri Jul 24, 2015 2:28 am
Location: Las Vegas, USA
Thanked: 19 times
Followed by:37 members

by Max@Math Revolution » Wed Aug 21, 2019 11:43 pm
=>

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) = 2/7, f(2^{1001}, 7) = 4/7, f(2^{1002}, 7) = 1/7, ..., and f(2^{2019}, 7) = 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) = (2/7 + 4/7 + 1/7)*(1020/3) = 340.

Therefore, D is the answer.
Answer: D
Top
Post new topic Post Reply
• Page 1 of 1