Forum Home > GMAT > Quantitative > Problem Solving (PS)
Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Dec 1410:00 AM EST
-12:00 PM EST
Think a 100% GMAT Verbal score is out of your reach? Target Test Prep will make you think again! Our course uses techniques such as topical study and spaced repetition to maximize knowledge retention and make studying simple and fun.
Dec 1411:00 AM IST
-01:00 PM IST
Attend this session to learn how to Deconstruct arguments, Pre-Think Author’s Assumptions, and apply Negation Test.
Dec 1310:00 AM EST
-12:00 PM EST
Have you ever wondered how to score a PERFECT 805 on the GMAT? Meet Julia, a banking professional who used the Target Test Prep course to achieve this incredible feat. Julia's story is nothing short of an inspiration.- Dec 15
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Dec 1608:00 AM PST
-11:59 PM PST
GMAT Club 12 Days of Christmas is a 4th Annual GMAT Club Winter Competition based on solving questions. This is the Winter GMAT competition on GMAT Club with an amazing opportunity to win over $40,000 worth of prizes!- Dec 17
09:00 AM PST
-10:00 AM PST
Join Manhattan Prep instructor Whitney Garner for a fun—and thorough—review of logic-based (non-math) problems, with a particular emphasis on Data Sufficiency and Two-Parts. - Dec 18
10:00 AM PST
-11:00 AM PST
Here is the essential guide to securing scholarships as an MBA student! In this video, we explore the various types of scholarships available, including need-based and merit-based options.
09 Jan 2017, 03:17
Kudos
Bookmarks
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
42% (02:17) correct
58%
(02:30)
wrong
based on 19
sessions
History
Date
Time
Result
Not Attempted Yet
If x and n are integers such that x = 1^1 + 2^2 + 3^3 + . . . + n^n, what is the remainder when x is divided by 5?
(1) n has a value of either 6 or 12.
(2) n is a multiple of 6.
(1) n has a value of either 6 or 12.
(2) n is a multiple of 6.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Data Sufficiency (DS) Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
09 Jan 2017, 07:14
Kudos
Bookmarks
Bunuel
If x and n are integers such that x = 1^1 + 2^2 + 3^3 + . . . + n^n, what is the remainder when x is divided by 5?
(1) n has a value of either 6 or 12.
(2) n is a multiple of 6.
(1) n has a value of either 6 or 12.
(2) n is a multiple of 6.
(1)
n=6
\(\frac{(1^1 + 2^2 + 3^3 + 4^4 + 5^5 + 6^6)}{5} = \frac{(1 + (-1) + 2 + (-1)^4 + 0 + 1^6)}{5} = \frac{4}{5}\) remainder 4
n=12
\(7^7 + 8^8 + 9^9 + 10^{10} + 11^{11} + 12^{12}\)
Last digits: \(\frac{(3 + 6 + 9 + 0 + 1 + 6)}{5} = \frac{5}{5}\) remainder 0 ---->Total remainder n=12 = 4 + 0 = 4
Sufficient
(2) We'll take next multiple of 6 - 18
\(13^{13} + 14^{14} + 15^{15} + 16^{16} + 17^{17} + 18^{18}\)
Last digits - 3 + 6 + 5 + 6 + 7 + 4 = 31 remainder 1
Total remainder n=18 4 + 1 = 5 ---> remainder 0.
We have remainders 4 and 0. Insufficient.
Unswer A
10 Jan 2017, 19:08
Kudos
Bookmarks
This one is E for me.
Way I went about this: count just the units digits of 1^1, 2^2...all the way up to 12^12 since statement 1 specifics n is either 6 or 12. Taking the first statement, if n=6, then we have:
1+4+7+6+5+6.....adds up to 29, remainder is 4.
On the other hand, if n=12, then we have:
1+4+7+6+5+6+3+6+9+0+1+2....adds up to 50. Remainder when divided by 5 is 0. Insufficient.
Statement 2: n is a multiple of 6. We just proved 6 and 12 will yield different remainders when divided by 5. Insufficient.
Both together: we don't learn anything new, 6 and 12 still fit and still yield different remainders.
E.
Way I went about this: count just the units digits of 1^1, 2^2...all the way up to 12^12 since statement 1 specifics n is either 6 or 12. Taking the first statement, if n=6, then we have:
1+4+7+6+5+6.....adds up to 29, remainder is 4.
On the other hand, if n=12, then we have:
1+4+7+6+5+6+3+6+9+0+1+2....adds up to 50. Remainder when divided by 5 is 0. Insufficient.
Statement 2: n is a multiple of 6. We just proved 6 and 12 will yield different remainders when divided by 5. Insufficient.
Both together: we don't learn anything new, 6 and 12 still fit and still yield different remainders.
E.
