Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 47157

Question Stats:
70% (01:40) correct 30% (01:41) wrong based on 98 sessions
HideShow timer Statistics



Math Expert
Joined: 02 Sep 2009
Posts: 47157

Re M2826
[#permalink]
Show Tags
16 Sep 2014, 01:29
Official Solution:If \(n\) is a nonnegative integer and the remainder when \(3^n\) is divided by 4 is a multiple of 3, then which of the following must be true? I. \(n^2\) divided by 4 yields the reminder of 1 II. \((2)^n\) is less than 0 III. \(n\) is a prime number A. \(I\) only B. \(II\) only C. \(III\) only D. \(I\) and \(II\) only E. \(II\) and \(III\) only \(3^0=1\).The remainder when 1 is divided by 4 is 1; \(3^1=3\). The remainder when 3 is divided by 4 is 3; \(3^2=9\). The remainder when 9 is divided by 4 is 1; \(3^3=27\). The remainder when 27 is divided by 4 is 3; ... We can see that in order the condition to hold true \(n\) must be odd. I. \(n^2\) divided by 4 yields the reminder of 1. Since \(odd^2\) divided by 4 always yields the reminder of 1, then this statement must be true. II. \((2)^n\) is less than 0. Since \((2)^{odd} \lt 0\), then this statement must be true. III. \(n\) is a prime number. Not necessarily true. Answer: D
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 14 Jan 2012
Posts: 13

Re: M2826
[#permalink]
Show Tags
20 Sep 2014, 00:45
My apologies for a simple question, but how is it possible to prove that odd^2 divided by 4 always yields the reminder of 1?
Is it correct based on (o/2)*(o/2)=(q*2+1)(q*2+1), so after multiplication we can conclude that R=1
Or based on (odd^21+1)/4, where odd^21 is always divisible by 4
Or is it a rule that i missed?
My concern aroused cause originally i came to the conclusion base on several numbers substitution.



Math Expert
Joined: 02 Sep 2009
Posts: 47157

Re: M2826
[#permalink]
Show Tags
20 Sep 2014, 13:27
Boycot wrote: My apologies for a simple question, but how is it possible to prove that odd^2 divided by 4 always yields the reminder of 1?
Is it correct based on (o/2)*(o/2)=(q*2+1)(q*2+1), so after multiplication we can conclude that R=1
Or based on (odd^21+1)/4, where odd^21 is always divisible by 4
Or is it a rule that i missed?
My concern aroused cause originally i came to the conclusion base on several numbers substitution. An odd number can be represented as 2k + 1 > (2k + 1)^2 = 4k^2 + 4k + 1 > first two terms are divisible by 4, so the remainder would come only from the third term > 1 divided by 4 yields the remainder of 1. Check it with numbers: 1^2 = 1 yields the remainder of 1 when divided by 4; 3^2 = 9 yields the remainder of 1 when divided by 4; 5^2 = 25 yields the remainder of 1 when divided by 4; ... Hope it's clear.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 24 Jun 2015
Posts: 46

Re: M2826
[#permalink]
Show Tags
02 Jul 2015, 05:48
Bunuel wrote: Official Solution:
If \(n\) is a nonnegative integer and the remainder when \(3^n\) is divided by 4 is a multiple of 3, then which of the following must be true? I. \(n^2\) divided by 4 yields the reminder of 1 II. \((2)^n\) is less than 0 III. \(n\) is a prime number
A. \(I\) only B. \(II\) only C. \(III\) only D. \(I\) and \(II\) only E. \(II\) and \(III\) only
\(3^0=1\).The remainder when 1 is divided by 4 is 1; \(3^1=3\). The remainder when 3 is divided by 4 is 3; \(3^2=9\). The remainder when 9 is divided by 4 is 1; \(3^3=27\). The remainder when 27 is divided by 4 is 3; ... We can see that in order the condition to hold true \(n\) must be odd. I. \(n^2\) divided by 4 yields the reminder of 1. Since \(odd^2\) divided by 4 always yields the reminder of 1, then this statement must be true. II. \((2)^n\) is less than 0. Since \((2)^{odd} \lt 0\), then this statement must be true. III. \(n\) is a prime number. Not necessarily true.
Answer: D Hi Bunuel, The first conclusion made was that N must be odd because of the pattern of 3^n, but is there another easy way to find this conclusion? My doubt is it is neccessary to do the numbers in order to see a pattern... Thanks a lot. Regards. Luis Navarro Looking for 700



Math Expert
Joined: 02 Sep 2009
Posts: 47157

luisnavarro wrote: Bunuel wrote: Official Solution:
If \(n\) is a nonnegative integer and the remainder when \(3^n\) is divided by 4 is a multiple of 3, then which of the following must be true? I. \(n^2\) divided by 4 yields the reminder of 1 II. \((2)^n\) is less than 0 III. \(n\) is a prime number
A. \(I\) only B. \(II\) only C. \(III\) only D. \(I\) and \(II\) only E. \(II\) and \(III\) only
\(3^0=1\).The remainder when 1 is divided by 4 is 1; \(3^1=3\). The remainder when 3 is divided by 4 is 3; \(3^2=9\). The remainder when 9 is divided by 4 is 1; \(3^3=27\). The remainder when 27 is divided by 4 is 3; ... We can see that in order the condition to hold true \(n\) must be odd. I. \(n^2\) divided by 4 yields the reminder of 1. Since \(odd^2\) divided by 4 always yields the reminder of 1, then this statement must be true. II. \((2)^n\) is less than 0. Since \((2)^{odd} \lt 0\), then this statement must be true. III. \(n\) is a prime number. Not necessarily true.
Answer: D Hi Bunuel, The first conclusion made was that N must be odd because of the pattern of 3^n, but is there another easy way to find this conclusion? My doubt is it is neccessary to do the numbers in order to see a pattern... Thanks a lot. Regards. Luis Navarro Looking for 700 Yes, it's the easiest ad not hard at all way. Check Units digits, exponents, remainders problems directory in our Special Questions Directory.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 24 Jun 2015
Posts: 46

Re: M2826
[#permalink]
Show Tags
02 Jul 2015, 06:23
Thanks¡¡



Intern
Joined: 17 Nov 2016
Posts: 1

Re M2826
[#permalink]
Show Tags
26 Nov 2016, 15:30
I think this is a highquality question and the explanation isn't clear enough, please elaborate. II. (−2)n=1 when n=0?



Math Expert
Joined: 02 Sep 2009
Posts: 47157

Re: M2826
[#permalink]
Show Tags
27 Nov 2016, 01:44



Intern
Joined: 10 Dec 2016
Posts: 3

Re: M2826
[#permalink]
Show Tags
25 Dec 2016, 12:08
If we take ,for example, n=6: We get 3^6= 4*180 + 9 ( the remainder 9 is a multiple of 3 ). And yet for n=6, none of the three conditions has to be true:  6²= 36. and dividing 36 by 4 doesn't yield a remainder of 1.  (2)^6 is positive.  6 is not a prime number. Can someone please tell me where I'm mistaken ? because for me, none of the conditions has to be true, thus none of the answers is correct.



Math Expert
Joined: 02 Sep 2009
Posts: 47157

Re: M2826
[#permalink]
Show Tags
26 Dec 2016, 10:47
LamyaaBezoui wrote: If we take ,for example, n=6: We get 3^6= 4*180 + 9 ( the remainder 9 is a multiple of 3 ). And yet for n=6, none of the three conditions has to be true:  6²= 36. and dividing 36 by 4 doesn't yield a remainder of 1.  (2)^6 is positive.  6 is not a prime number. Can someone please tell me where I'm mistaken ? because for me, none of the conditions has to be true, thus none of the answers is correct. The remainder of 3^6/4 is 1 not 9. The reminder upon division by 4 cannot possible be 9 it can be 0, 1, 2, or 3. The remainder is always less than the divisor.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 27 Oct 2015
Posts: 19

Re: M2826
[#permalink]
Show Tags
01 Jan 2017, 03:45
Hu Buenel,
II. (−2)^n is less than 0.
What about when n = 0?
This doesnt satisfy.
Please help



Math Expert
Joined: 02 Sep 2009
Posts: 47157

Re: M2826
[#permalink]
Show Tags
01 Jan 2017, 08:16



Intern
Joined: 27 Aug 2016
Posts: 3

Re M2826
[#permalink]
Show Tags
25 Oct 2017, 20:13
But for the case, when n =0, which hold true for original condition, then choice I does not work. 0 when divided by 4 leaves 4 as remainder. Am I missing sth ?



Math Expert
Joined: 02 Sep 2009
Posts: 47157

Re: M2826
[#permalink]
Show Tags
25 Oct 2017, 22:33










