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

 It is currently 22 Oct 2019, 14:46

### 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

# S98-01

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 58427

### Show Tags

16 Sep 2014, 01:52
00:00

Difficulty:

75% (hard)

Question Stats:

41% (01:33) correct 59% (01:44) wrong based on 29 sessions

### HideShow timer Statistics

What is the remainder when you divide $$2^{200}$$ by 7?

A. 1
B. 2
C. 3
D. 4
E. 5

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 58427

### Show Tags

16 Sep 2014, 01:52
Official Solution:

What is the remainder when you divide $$2^{200}$$ by 7?

A. 1
B. 2
C. 3
D. 4
E. 5

We cannot compute $$2^{200}$$ in anywhere near the time allotted, so we should look for a pattern in much simpler problems that we can scale up to $$2^{200}$$.

The simpler problems we should solve are these:

What is the remainder when you divide 2 by 7?

What is the remainder when you divide $$2^2$$ by 7?

What is the remainder when you divide $$2^3$$ by 7?

What is the remainder when you divide $$2^4$$ by 7?

... and so on with the powers of 2.

2 divided by 7 leaves remainder 2.

$$2^2$$ (which equals 4) divided by 7 leaves remainder 4.

$$2^3$$ (which equals 8) divided by 7 leaves remainder 1.

$$2^4$$ (which equals 16) divided by 7 leaves remainder 2.

$$2^5$$ (which equals 32) divided by 7 leaves remainder 4.

$$2^6$$ (which equals 64) divided by 7 leaves remainder 1.

We should stop as soon as we notice that the cycle will repeat itself forever in this pattern: [2, 4, 1]. Every third remainder is the same. (From here on out, "remainder" always means "remainder after we divide by 7.") Since every third remainder is the same, we should look at the remainder when the power is a multiple of 3. The remainders of 23 and 26 are 1. Thus, the remainder of 2 raised to a power that is any multiple of 3 is 1.

Now, 200 is not a multiple of 3, but we can look for a multiple of 3 near 200. 201 is a multiple of 3 (its digits add to 3), so $$2^{201}$$ has a remainder of 1. Finally, we notice that the remainder of $$2^{200}$$ must be one position earlier in the cycle than the remainder of $$2^{201}$$. Since the cycle is [2, 4, 1], the remainder of $$2^{200}$$ is 4.

_________________
Intern
Joined: 20 Aug 2016
Posts: 43
GMAT 1: 570 Q46 V23
GMAT 2: 610 Q49 V25
GMAT 3: 620 Q45 V31
WE: Information Technology (Other)

### Show Tags

04 Oct 2018, 06:27
1
2^5 /7 gives 4 as remainder. Hence, power 200, which is multiple of 5, will give remainder as 4, as well
Intern
Joined: 03 Jul 2018
Posts: 5

### Show Tags

21 Mar 2019, 04:56
Saps

That's a wrong deduction. Since 200 is a multiple of 2^1 , 2^2, 2^3 ....etc .

Posted from my mobile device
Intern
Joined: 03 Jul 2018
Posts: 5

### Show Tags

21 Mar 2019, 05:11
Bunuel,

Can you please explain the last part of the answer - "Now, 200 is not a multiple of 3, but we can look for a multiple of 3 near 200. 201 is a multiple of 3 (its digits add to 3), so 22012201 has a remainder of 1. Finally, we notice that the remainder of 22002200 must be one position earlier in the cycle than the remainder of 22012201. Since the cycle is [2, 4, 1], the remainder of 22002200 is 4."

I got stuck as I chose 2^198 as the closest no divisible by 3, then 200 is two positions after in the cycle (2,4,1) and so the remainder is 1 ??!

Posted from my mobile device
Math Expert
Joined: 02 Sep 2009
Posts: 58427

### Show Tags

21 Mar 2019, 05:13
Geetshri wrote:
Bunuel,

Can you please explain the last part of the answer - "Now, 200 is not a multiple of 3, but we can look for a multiple of 3 near 200. 201 is a multiple of 3 (its digits add to 3), so 22012201 has a remainder of 1. Finally, we notice that the remainder of 22002200 must be one position earlier in the cycle than the remainder of 22012201. Since the cycle is [2, 4, 1], the remainder of 22002200 is 4."

I got stuck as I chose 2^198 as the closest no divisible by 3, then 200 is two positions after in the cycle (2,4,1) and so the remainder is 1 ??!

Posted from my mobile device

Check here: https://gmatclub.com/forum/what-is-the- ... 40821.html
_________________
Intern
Joined: 08 Apr 2019
Posts: 2
Location: India
Concentration: Leadership, General Management

### Show Tags

13 Apr 2019, 02:23
I agree with explanation.
Intern
Joined: 28 Nov 2018
Posts: 5

### Show Tags

16 Apr 2019, 23:54
how can 2 divided by 7 = 2 remainder?
Math Expert
Joined: 02 Sep 2009
Posts: 58427

### Show Tags

17 Apr 2019, 00:01
7sharma wrote:
how can 2 divided by 7 = 2 remainder?

Let me ask you a question: how many leftover apples would you have if you had 1 apple and wanted to distribute in 9 baskets evenly? Each basket would get 0 apples and 1 apple would be leftover (remainder).

When a divisor is more than dividend, then the remainder equals to the dividend, for example:
3 divided by 4 yields the reminder of 3: $$3=4*0+3$$;
9 divided by 14 yields the reminder of 9: $$9=14*0+9$$;
1 divided by 9 yields the reminder of 1: $$1=9*0+1$$.

6. Remainders

For more:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

Hope it helps.
_________________
Re: S98-01   [#permalink] 17 Apr 2019, 00:01
Display posts from previous: Sort by

# S98-01

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne