Last visit was: 13 Jul 2024, 21:29 It is currently 13 Jul 2024, 21:29
Close
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
Your Progress

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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
SORT BY:
Date
Tags:
Show Tags
Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 94341
Own Kudos [?]: 640561 [18]
Given Kudos: 85005
Send PM
Math Expert
Joined: 02 Sep 2009
Posts: 94341
Own Kudos [?]: 640561 [0]
Given Kudos: 85005
Send PM
RC & DI Moderator
Joined: 02 Aug 2009
Status:Math and DI Expert
Posts: 11468
Own Kudos [?]: 34285 [0]
Given Kudos: 322
Send PM
avatar
Intern
Intern
Joined: 22 Mar 2015
Posts: 3
Own Kudos [?]: 3 [0]
Given Kudos: 0
Send PM
Re: A Tricky Question on Negative Remainders [#permalink]
Hi everyone,

I am new and I was just reading the post but I was somehow lost when following the solution...so I tried to solve myself and I did in this way:

I wrote down the following in order to find a possible pattern (R=remainder):

k | 3^k | R when divided by 5 |
0 | 1___|_______1________|
1 | 3___|_______3________|
2 | 9___|_______4________|
3 | 27__|_______2________|
4 | 81__|_______1________|
5 | 243_|_______3________|

Stop! OK I found a general pattern for the remainder cyclical (1,3,4,2)...then since 3^(7^11) is equal to 3^77 then is sufficient to look at the line where k is 7 (because 77 is eleven times 7)...that is the same as the line with k=3 so R=2! Indeed...

6 | not needed |_______4________|
7 | not needed |_______2________|

So the answer is 2 that is (C).

Am I wrong? It was quite fast in this way...

Thanks to everybody!

Padener

P.s.
Sorry for the _____ but I did not know how to format properly the table....
Math Expert
Joined: 02 Sep 2009
Posts: 94341
Own Kudos [?]: 640561 [0]
Given Kudos: 85005
Send PM
Re: A Tricky Question on Negative Remainders [#permalink]
Expert Reply
padener wrote:
Hi everyone,

I am new and I was just reading the post but I was somehow lost when following the solution...so I tried to solve myself and I did in this way:

I wrote down the following in order to find a possible pattern (R=remainder):

k | 3^k | R when divided by 5 |
0 | 1___|_______1________|
1 | 3___|_______3________|
2 | 9___|_______4________|
3 | 27__|_______2________|
4 | 81__|_______1________|
5 | 243_|_______3________|

Stop! OK I found a general pattern for the remainder cyclical (1,3,4,2)...then since 3^(7^11) is equal to 3^77 then is sufficient to look at the line where k is 7 (because 77 is eleven times 7)...that is the same as the line with k=3 so R=2! Indeed...

6 | not needed |_______4________|
7 | not needed |_______2________|

So the answer is 2 that is (C).

Am I wrong? It was quite fast in this way...

Thanks to everybody!

Padener

P.s.
Sorry for the _____ but I did not know how to format properly the table....


When getting the pattern of the units digit of x^k you should always start with k=1, not k=0.
avatar
Intern
Intern
Joined: 22 Mar 2015
Posts: 3
Own Kudos [?]: 3 [0]
Given Kudos: 0
Send PM
Re: A Tricky Question on Negative Remainders [#permalink]
Yes you are right because k=0 will always give 1 as results...so no meaning.

But apart from that is right the approach?

Thanks for the answer.

Padener
Math Expert
Joined: 02 Sep 2009
Posts: 94341
Own Kudos [?]: 640561 [0]
Given Kudos: 85005
Send PM
Re: A Tricky Question on Negative Remainders [#permalink]
Expert Reply
Bunuel wrote:
padener wrote:
Hi everyone,

I am new and I was just reading the post but I was somehow lost when following the solution...so I tried to solve myself and I did in this way:

I wrote down the following in order to find a possible pattern (R=remainder):

k | 3^k | R when divided by 5 |
0 | 1___|_______1________|
1 | 3___|_______3________|
2 | 9___|_______4________|
3 | 27__|_______2________|
4 | 81__|_______1________|
5 | 243_|_______3________|

Stop! OK I found a general pattern for the remainder cyclical (1,3,4,2)...then since 3^(7^11) is equal to 3^77 then is sufficient to look at the line where k is 7 (because 77 is eleven times 7)...that is the same as the line with k=3 so R=2! Indeed...

6 | not needed |_______4________|
7 | not needed |_______2________|

So the answer is 2 that is (C).

Am I wrong? It was quite fast in this way...

Thanks to everybody!

Padener

P.s.
Sorry for the _____ but I did not know how to format properly the table....


When getting the pattern of the units digit of x^k you should always start with k=1, not k=0.


The red part is not correct: 3^7^11 = 3^77 but 3^(7^11) = 3 to the power of (7^11).

Theory on Exponents: math-number-theory-88376.html
Tips on Exponents: exponents-and-roots-on-the-gmat-tips-and-hints-174993.html

Check Units digits, exponents, remainders problems directory in our Special Questions Directory.
Intern
Intern
Joined: 22 Apr 2019
Posts: 26
Own Kudos [?]: 8 [0]
Given Kudos: 13
Send PM
Re: A Tricky Question on Negative Remainders [#permalink]
Is there something wrong in my methodology?

7^11 .. 7 has cycle of 7, 9, 3, 1.. so the remainder is 3.
3^3 = 27
27 / 5 = remainder of 2
GMAT Tutor
Joined: 24 Jun 2008
Posts: 4128
Own Kudos [?]: 9437 [0]
Given Kudos: 91
 Q51  V47
Send PM
Re: A Tricky Question on Negative Remainders [#permalink]
Expert Reply
Tactic101 wrote:
Is there something wrong in my methodology?

7^11 .. 7 has cycle of 7, 9, 3, 1.. so the remainder is 3.


When you divide 3^k by 5 or 10, and want to know the remainder, then you want to know where you'll be in the units digit pattern of 3^k. Because the units digit pattern of 3^k repeats in 'blocks' (cycles) of 4 terms, you need to know the remainder when you divide k by 4 in order to know where you'll be in that pattern. In this question, k = 7^11, so you'd need to find the remainder when you divide 7^11 by 4. What you found above is the remainder when you divide 7^11 by 10. So you correctly determined that 7^11 has a units digit of 3, but that alone doesn't tell you the remainder when you divide 7^11 by 4 (since some numbers that end in 3 will give a remainder of 1 when you divide by 4, and some will give a remainder of 3 when you divide by 4). It's because it's not straightforward to find the remainder when you divide 7^11 by 4 that the posters above needed some lengthier solutions.
Director
Director
Joined: 21 Feb 2017
Posts: 509
Own Kudos [?]: 1074 [0]
Given Kudos: 1091
Location: India
GMAT 1: 700 Q47 V39
Send PM
Re: A Tricky Question on Negative Remainders [#permalink]
Tactic101 wrote:
Is there something wrong in my methodology?

7^11 .. 7 has cycle of 7, 9, 3, 1.. so the remainder is 3.
3^3 = 27
27 / 5 = remainder of 2


Hi
"7^11 .. 7 has cycle of 7, 9, 3, 1.. so the remainder is 3". Here the units digit would be 3 and not the remainder. Am I correct?
how to solve ahead if we know 3^m3/5? could someone help pls.
User avatar
Non-Human User
Joined: 09 Sep 2013
Posts: 33964
Own Kudos [?]: 851 [0]
Given Kudos: 0
Send PM
Re: A Tricky Question on Negative Remainders [#permalink]
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.
GMAT Club Bot
Re: A Tricky Question on Negative Remainders [#permalink]
Moderator:
Math Expert
94341 posts