GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 15 Oct 2019, 19:22

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Find the remainder when the sum of 2^100 + 2^200 + 2^300+

Author Message
TAGS:

### Hide Tags

Manager
Joined: 15 Feb 2016
Posts: 59
GMAT 1: 710 Q48 V40
Re: Find the remainder when the sum of 2^100 + 2^200 + 2^300+  [#permalink]

### Show Tags

15 Feb 2016, 01:12
1
mindmind wrote:
Find the remainder when the sum of $$2^{100} + 2^ {200} + 2^ {300}+........+ 2^ {10000}$$ is divided by 7

A. 0
B. 2
C. 1
D. 6
E. 5

Solved this in one minute.

Just remember to use the concept explained @ quarter-wit-quarter-wisdom-a-remainders-post-for-the-geek-in-you and half of the remainder Qs become a piece of cake

All the terms in the expression (a+b)^z are divisible by a except b^z

(8=7+1) Make the modifications in the expression so that it is in some relation with the given divisor (usually plus or minus 1)

2^100 = 2* 8^33 = 2* (7+1)^33
2^200 = 4* 8^66 = 4* (7+1)^66
2^300 = 1* 8^100 = 1* (7+1)^100
2^400 = 2* 8^133 = 2* (7+1)^133
2^500 = 4* 8^166 = 4* (7+1)^166
2^600 = 1* 8^200 = 1* (7+1)^200
Observer that this it is a repeat cycle of 2, 4 and 1

Make a note of total number of terms = 100
remainder from first term = 2*1
remainder from second term = 4*1
remainder from third term = 1*1

Sum of first three remainders = 7. Remainder =0 .Sum of first 99 remainders = 7*33 Remainder =0 . Remainder on the 100th term = 2 (from the cycle)

Though I sound pretty confident I need someone to validate the method Thanks!
_________________
It is not who I am underneath but what I do that defines me.
Intern
Joined: 02 Feb 2016
Posts: 15
Re: Find the remainder when the sum of 2^100 + 2^200 + 2^300+  [#permalink]

### Show Tags

04 Mar 2016, 14:18
jns wrote:
mindmind wrote:
Find the remainder when the sum of $$2^{100} + 2^ {200} + 2^ {300}+........+ 2^ {10000}$$ is divided by 7

2^0 = 1. 1/7 => R=1, R = remainder
2^1 = 2. 2/7 => R=2
2^2 = 4. 4/7 => R=4
2^3 = 8. 8/7 => R=1
2^4 = 16. 16/7 => R=2 .... and the cycle repeats.

We know that the cycle repeats after every 3rd term - 1,2,4,1,2,4.... So, divide the power by 3 to find the number of complete cycles and the remaining powers.
2^100 = (2^99)*2 = R1*2 = 2. 2/7 => R=2 ------- (99/3 complete cycles and one 2 left.)
2^200 = (2^198)*(2^2) = R1*4 = 4. /7 => R=4 ------- (198/3 complete cycles and two 2 left.)
2^300 = R1. 1/7 => R=1 ------- (100 complete cycles and no 2 left.)
Similarly, 2^400 => R=2, 2^500 => R=4, 2^600 => R=1 and the cycle repeats. 2^10000 = (2^9999)*2 => R2.

So, $$2^{100} + 2^ {200} + 2^ {300}+........+ 2^ {10000}$$ in terms of remainders is equal to 2 + 4 + 1 + 2 + 4 + 1 +....... 2
Number of terms in the sequence = 10000/100 = 100 = 99 + 1. Each remainder cycle consists of 3 digits - 2,4,1. There are 99/3 = 33 complete cycles and 1 left out 100th term
Therefore, 2 + 4 + 1 + 2 + 4 + 1 +....... 2 = (2+4+1)*33 + 2 = > 7*33 + 2. On dividing this result by 7, the remainder is 2.

Why do you start the cycle (2^0; 2^1; 2^2 etc) from 2^0? Is it a special case? I've always started from n^1 and the answer was right. Thank you!
Math Expert
Joined: 02 Aug 2009
Posts: 7960
Re: Find the remainder when the sum of 2^100 + 2^200 + 2^300+  [#permalink]

### Show Tags

05 Mar 2016, 04:41
1
Viktoriaa wrote:
jns wrote:
mindmind wrote:
Find the remainder when the sum of $$2^{100} + 2^ {200} + 2^ {300}+........+ 2^ {10000}$$ is divided by 7

2^0 = 1. 1/7 => R=1, R = remainder
2^1 = 2. 2/7 => R=2
2^2 = 4. 4/7 => R=4
2^3 = 8. 8/7 => R=1
2^4 = 16. 16/7 => R=2 .... and the cycle repeats.

We know that the cycle repeats after every 3rd term - 1,2,4,1,2,4.... So, divide the power by 3 to find the number of complete cycles and the remaining powers.
2^100 = (2^99)*2 = R1*2 = 2. 2/7 => R=2 ------- (99/3 complete cycles and one 2 left.)
2^200 = (2^198)*(2^2) = R1*4 = 4. /7 => R=4 ------- (198/3 complete cycles and two 2 left.)
2^300 = R1. 1/7 => R=1 ------- (100 complete cycles and no 2 left.)
Similarly, 2^400 => R=2, 2^500 => R=4, 2^600 => R=1 and the cycle repeats. 2^10000 = (2^9999)*2 => R2.

So, $$2^{100} + 2^ {200} + 2^ {300}+........+ 2^ {10000}$$ in terms of remainders is equal to 2 + 4 + 1 + 2 + 4 + 1 +....... 2
Number of terms in the sequence = 10000/100 = 100 = 99 + 1. Each remainder cycle consists of 3 digits - 2,4,1. There are 99/3 = 33 complete cycles and 1 left out 100th term
Therefore, 2 + 4 + 1 + 2 + 4 + 1 +....... 2 = (2+4+1)*33 + 2 = > 7*33 + 2. On dividing this result by 7, the remainder is 2.

Why do you start the cycle (2^0; 2^1; 2^2 etc) from 2^0? Is it a special case? I've always started from n^1 and the answer was right. Thank you!

hi,
you are CORRECT in your approach.
although the person has calculated for 2^0, but has not used in his calculations for finding remainder of 2^100...
You should maintain the approach you have been following

_________________
Current Student
Joined: 12 Aug 2015
Posts: 2569
Schools: Boston U '20 (M)
GRE 1: Q169 V154
Re: Find the remainder when the sum of 2^100 + 2^200 + 2^300+  [#permalink]

### Show Tags

21 Mar 2016, 05:31
Bunuel
Need Some Input..
I solved the Question using this approach=>
there are 100 terms in the series.
First term =2^100 => 2*[7+1]^33 => using the Binomial expansion => 2* [7K +1] = 7p+2
2nd term => 2^200 => 4*[7+1]^66 => 7p+4
3rd term => 2^300 => 7p+1
and the next terms are all following the same pattern ..
hence the series can be written as => 7p+2 + 7t+4 + 7s+1 +....
now i noticed that there are 100 terms => sum of the first 99 terms => 7c for some c
hence the last term = 7r+2 => sum of all the terms => 7z+2 hence 2 is the remainder..
I look forward to a method other than this too..

also is this approach right ?
thanks and regards
_________________
Senior Manager
Joined: 13 Oct 2016
Posts: 359
GPA: 3.98
Re: Find the remainder when the sum of 2^100 + 2^200 + 2^300+  [#permalink]

### Show Tags

01 Dec 2016, 07:17
1
mindmind wrote:
Find the remainder when the sum of $$2^{100} + 2^ {200} + 2^ {300}+........+ 2^ {10000}$$ is divided by 7

A. 0
B. 2
C. 1
D. 6
E. 5

Cyclicity is a good approach but modular arithmetic will help us to solve this question a little bit faster.

We have:

$$\frac{2^{100} + 2^{200} + 2^{300} + 2^{400} + … + 2^{10000}}{7}$$

$$2^3 = 1 (mod _7)$$

$$2^{100} = 2^{99}*2 = (2^3)^{33}*2 = 1^{33}*2 = 2$$

$$2^{100} + 2^{200} + 2^{300} + 2^{400} + … + 2^{10000} = (2^{100})^1 + (2^{100})^2 + (2^{100})^3 + … + (2^{100})^{100}$$

In mod 7 this is the same as:

$$\frac{2 + 2^2 + 2^3 + 2^4 + 2^5 + … + 2^{100}}{7}$$ where the nominator is the sum of geometric progression.

$$\frac{2*(2^{100}-1)}{(2-1)*7} = \frac{2*(2^{100}-1)}{7}$$

But we already know that $$2^{100} = 2 (mod_7)$$

And our expression boils down to:

$$\frac{2*(2-1)}{7} = \frac{2}{7}$$

Our remainder is $$2$$.

Senior Manager
Joined: 27 Dec 2016
Posts: 309
Re: Find the remainder when the sum of 2^100 + 2^200 + 2^300+  [#permalink]

### Show Tags

24 Dec 2017, 20:14
chetan2u Bunuel niks18 VeritasPrepKarishma

Hi,
I was wondering if one of you experts could please check my work? I was able to get to the right answer in under 2 mins but I am not entirely positive if my way is correct. I simply took the first term and last term and added them together to get the final equation as:

(2^100+2^10000)/7 - (2^100(1*2^9900))/7 - 2^100/7= Remainder of 2 & 2^9900/7= Remainder of 1. I then multiplied the remainders together (2*1)/7 to finally get a remainder of 2.

Please let me know if my way was correct? Would greatly appreciate it!
Retired Moderator
Joined: 25 Feb 2013
Posts: 1179
Location: India
GPA: 3.82
Find the remainder when the sum of 2^100 + 2^200 + 2^300+  [#permalink]

### Show Tags

24 Dec 2017, 21:32
csaluja wrote:
chetan2u Bunuel niks18 VeritasPrepKarishma

Hi,
I was wondering if one of you experts could please check my work? I was able to get to the right answer in under 2 mins but I am not entirely positive if my way is correct. I simply took the first term and last term and added them together to get the final equation as:

(2^100+2^10000)/7 - (2^100(1*2^9900))/7 - 2^100/7= Remainder of 2 & 2^9900/7= Remainder of 1. I then multiplied the remainders together (2*1)/7 to finally get a remainder of 2.

Please let me know if my way was correct? Would greatly appreciate it!

Hi csaluja,

At first I could not understand your logic for adding the two extremes. Also the highlighted portion is incorrect. if you take 2^100 out then the expression will be 2^100(1+something)

Note that this is a GP(geometric series) and formula to find sum of GP series $$=\frac{a_1(r^n-1)}{(r-1)}$$, where $$a_1=$$first term, $$r=$$ common ration$$=\frac{a_2}{a_1}$$ and $$n=$$number of terms

Also note that $$\frac{2^{100}}{7}=\frac{(2^3)^{33}*2}{7}=\frac{8^{33}*2}{7}$$, Now 8 leaves a remainder of 1 when divided by 7 and 2 leaves a remainder of 2

$$=>remainder = 1^{33}*2=2$$

Similarly $$2^{200}=(2^{100})^2$$, remainder for this term will be $$(2)^2$$

So we will have new G.P series of REMAINDERS

$$2 + 2^2 + 2^3 +......+ 2^{100}$$

Sum of the remainders: $$S_n = \frac{a(r^n - 1)}{r-1}$$, where $$a=2; r=2, n=100$$

$$S_{100}=\frac{2(2^{100} - 1)}{(2-1)} =>2(2^{100} - 1)=2^{101}-2$$

Hence $$Remainder = \frac{(2^{101}-2)}{7}=\frac{[(2^3)^{33}*2^2]}{7}-\frac{2}{7}$$

or, $$remainder= \frac{8^{33}*4}{7}-\frac{2}{7} =1^{33}*4-2=4-2=2$$
Non-Human User
Joined: 09 Sep 2013
Posts: 13168
Re: Find the remainder when the sum of 2^100 + 2^200 + 2^300+  [#permalink]

### Show Tags

11 Feb 2019, 23:35
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: Find the remainder when the sum of 2^100 + 2^200 + 2^300+   [#permalink] 11 Feb 2019, 23:35

Go to page   Previous    1   2   [ 28 posts ]

Display posts from previous: Sort by