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gurpreetsingh
Joined: 12 Oct 2009
Last visit: 15 Jun 2019
Posts: 2,266
Given Kudos: 235
Status:<strong>Nothing comes easy: neither do I want.</strong>
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Products:
03 Sep 2010, 20:39
Kudos
Bookmarks
Hi all,
I will post weekly question on Saturday's and the Solution on Monday's 22:00 IST. Pm me if I forget .
Lets not discuss whether this type of question can come on Gmat or not rather learn something new from the question. Even if you are not able to solve it in 2mins just try to crack some logic. I m sure you guys will learn something new every week.
Strictly No replies here. This thread will be updated with question and its solution separately so that it can be revised on one page.
All the replies on this thread will be deleted and warning will be issued. Kindly be supportive.
-----------------------NO Discussion------------------------
Disclaimer : Questions not created by me , but are from multiple mixed sources. The credit goes to the creator.
I will post weekly question on Saturday's and the Solution on Monday's 22:00 IST. Pm me if I forget .
Lets not discuss whether this type of question can come on Gmat or not rather learn something new from the question. Even if you are not able to solve it in 2mins just try to crack some logic. I m sure you guys will learn something new every week.
Strictly No replies here. This thread will be updated with question and its solution separately so that it can be revised on one page.
All the replies on this thread will be deleted and warning will be issued. Kindly be supportive.
-----------------------NO Discussion------------------------
Disclaimer : Questions not created by me , but are from multiple mixed sources. The credit goes to the creator.
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gurpreetsingh
Joined: 12 Oct 2009
Last visit: 15 Jun 2019
Posts: 2,266
Own Kudos:
Given Kudos: 235
Status:<strong>Nothing comes easy: neither do I want.</strong>
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Products:
03 Sep 2010, 20:52
Kudos
Bookmarks
04 Sept 2010
Question : 1
Q1. Find the remainder when \(1*2 + 2*3 + 3*4 + 4*5 ............99*100\) is divided by 101.
A) 0
B) 1
C) 100
D) 99
E) 2
Discussion: happy-weekend-questions-04-sept-100375.html
Question : 1
Q1. Find the remainder when \(1*2 + 2*3 + 3*4 + 4*5 ............99*100\) is divided by 101.
A) 0
B) 1
C) 100
D) 99
E) 2
Discussion: happy-weekend-questions-04-sept-100375.html
Show Spoilersolution: 1
Sn = \(1*2 + 2*3 + 3*4 + 4*5 ............99*100\)
= summation of n*(n+1) where n varies from 1 to 99
= summation of n^2 + n = \(n*(n+1)*\frac{(2n+1)}{6}\) \(+ n*\frac{(n+1)}{2}\)
= \(n*(n+1)*\frac{(n+2)}{3}\)
Now put n =99 we get sum = \(99*100*\frac{101}{3}\)
This is divisible by 101, hence remainder is 0.
Answer is A.
= summation of n*(n+1) where n varies from 1 to 99
= summation of n^2 + n = \(n*(n+1)*\frac{(2n+1)}{6}\) \(+ n*\frac{(n+1)}{2}\)
= \(n*(n+1)*\frac{(n+2)}{3}\)
Now put n =99 we get sum = \(99*100*\frac{101}{3}\)
This is divisible by 101, hence remainder is 0.
Answer is A.
gurpreetsingh
Joined: 12 Oct 2009
Last visit: 15 Jun 2019
Posts: 2,266
Own Kudos:
Given Kudos: 235
Status:<strong>Nothing comes easy: neither do I want.</strong>
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Products:
11 Sep 2010, 11:45
Kudos
Bookmarks
11 Sept 2010
Question : 2
Q2. Find the remainder when \(1^{39} +2^{39} + 3^{39}..........12^{39}\) is divided by 39.
A) 0
B) 1
C) 3
D) 2
E) 12
Discussion: happy-weekend-question-11-sept-100862.html
Question : 2
Q2. Find the remainder when \(1^{39} +2^{39} + 3^{39}..........12^{39}\) is divided by 39.
A) 0
B) 1
C) 3
D) 2
E) 12
Discussion: happy-weekend-question-11-sept-100862.html
Show Spoilersolution: 2
Using the property : \(a^n + b^n\) is always divisible by \(a+b\) if n is odd
\(1^{39} +2^{39} + 3^{39}..........12^{39}\)
=> \(1^{39} +12^{39} + 2^{39} + 11^{39} +3^{39} + 10^{39} + ......6^{39} + 7^{39}\) is divisible by
1+12 + 2+11 + 3+10 + 4+9 + 5+8 + 7+6 = 6*13 = 39*2
=> it is divisible by 13 and the remainder is 0.
Answer is A.
In general, \(a^n + b^n +c^n........ z^n\) is divisible by \(a+b+c......z\) if n is odd
\(1^{39} +2^{39} + 3^{39}..........12^{39}\)
=> \(1^{39} +12^{39} + 2^{39} + 11^{39} +3^{39} + 10^{39} + ......6^{39} + 7^{39}\) is divisible by
1+12 + 2+11 + 3+10 + 4+9 + 5+8 + 7+6 = 6*13 = 39*2
=> it is divisible by 13 and the remainder is 0.
Answer is A.
In general, \(a^n + b^n +c^n........ z^n\) is divisible by \(a+b+c......z\) if n is odd
