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If n is a positive integer, what is the remainder when

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If n is a positive integer, what is the remainder when [#permalink]

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19 Nov 2010, 08:40
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If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?

A. 0
B. 1
C. 2
D. 3
E. 4
[Reveal] Spoiler: OA

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19 Nov 2010, 08:49
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azule45 wrote:
If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?

A 0
B 1
C 2
D 3
E 4

It should be $$3^{8n+3}+2$$.

The units digit of 3 in positive integer power has cyclicity of 4 for the unis digit:
3^1 --> the units digit is 3;
3^2 --> the units digit is 9;
3^3 --> the units digit is 7;
3^4 --> the units digit is 1;
3^5 --> the units digit is 3 AGAIN;
...

So, the units digit repeats the following pattern {3-9-7-1}-{3-9-7-1}-.... $$3^{8n+3}$$ will have the same units digit as $$3^3$$, which is 7 (remainder when 8n+3 divided by cyclicity 4 is 3). Thus the last digit of $$3^{8n+3}+2$$ will be $$7+2=9$$. Any positive integer with the unis digit of 9 divided by 5 gives the remainder of 4.

Check Number theory chapter of Math Book for more: math-number-theory-88376.html

Hope it helps.
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19 Nov 2010, 08:59
ahhh okay. i followed the same pattern for the exponents. the problem made it seem like +2 was in the exponent. makes perfect sense now. much appreciated, Bunuel.
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19 Nov 2010, 10:42
Great explanation.
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20 Nov 2010, 02:17
put n=0, then we have (3^3)+2=29. now 29/5, we have r=4. is my short cut ok ???

counter arguments welcome instead of kudos !!!
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20 Nov 2010, 12:20
anandthiru wrote:
put n=0, then we have (3^3)+2=29. now 29/5, we have r=4. is my short cut ok ???

counter arguments welcome instead of kudos !!!

n is given to be a positive integer so it cannot be 0. That said, whatever will be the last digit of $$3^{(8n+3)} + 2$$ for n = positive integer, will be the last digit when n = 0. (We know cyclicity works for n = 0 too)
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Senior Manager Joined: 20 Apr 2010 Posts: 250 Location: Hyderabad WE 1: 4.6 years Exp IT prof Followers: 8 Kudos [?]: 25 [0], given: 51 Re: Exponents [#permalink] Show Tags 22 Nov 2010, 13:38 Answer should be E i.e. 4 remember 3 has a cyclicity of 4 =>3,9,*7,*1 3^(8n+3) +2 => for n=1 => 8*1+3 =>11 =>3(on dividing 11/4) for n=2 =>19=>3(on dividing 19/4) for n=3 =>27=>3(on dividing 27/4) for n=4 =>35=>3(on dividing 35/4) for n=5 =>43=>3(on dividing 43/4) for n=6 =>51=>3(on dividing 51/4) therefore 3^3 gives 7 in units place => 7+2 => 9 in all the cases hence on dividing 9/5 remainder is 4 _________________ I will give a Fight till the End "To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed." - Bernard Edmonds A person who is afraid of Failure can never succeed -- Amneet Padda Don't Forget to give the KUDOS Math Expert Joined: 02 Sep 2009 Posts: 36590 Followers: 7092 Kudos [?]: 93390 [0], given: 10557 Re: If n is a positive integer, what is the remainder when [#permalink] Show Tags 06 Jun 2013, 05:33 Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HERE Theory on remainders problems: remainders-144665.html All DS remainders problems to practice: search.php?search_id=tag&tag_id=198 All PS remainders problems to practice: search.php?search_id=tag&tag_id=199 _________________ Intern Joined: 09 Apr 2013 Posts: 32 Location: United States Concentration: Social Entrepreneurship, Strategy GMAT 1: 750 Q50 V41 GPA: 3.55 Followers: 1 Kudos [?]: 38 [3] , given: 0 Re: If n is a positive integer, what is the remainder when [#permalink] Show Tags 06 Jun 2013, 14:00 3 This post received KUDOS you can split it up to make it look simpler. if x = 1 (3^8)(3^3) + 2 3^8 = 9^4 = 9 x 9 x 9 x 9 = 81 x 81 (units digit is 1) 3^3 = 27 (units digit is 7) 1 x 7 + 2 = 9 9/5 = 1 r4 Answer is E Manager Joined: 15 Apr 2013 Posts: 86 Location: India Concentration: Finance, General Management Schools: ISB '15 WE: Account Management (Other) Followers: 1 Kudos [?]: 93 [0], given: 61 Re: Exponents [#permalink] Show Tags 06 Jun 2013, 14:20 Bunuel wrote: azule45 wrote: If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5? A 0 B 1 C 2 D 3 E 4 It should be $$3^{8n+3}+2$$. The units digit of 3 in positive integer power has cyclicity of 4 for the unis digit: 3^1 --> the units digit is 3; 3^2 --> the units digit is 9; 3^3 --> the units digit is 7; 3^4 --> the units digit is 1; 3^5 --> the units digit is 3 AGAIN; ... So, the units digit repeats the following pattern {3-9-7-1}-{3-9-7-1}-.... $$3^{8n+3}$$ will have the same units digit as $$3^3$$, which is 7 (remainder when 8n+3 divided by cyclicity 4 is 3). Thus the last digit of $$3^{8n+3}+2$$ will be $$7+2=9$$. Any positive integer with the unis digit of 9 divided by 5 gives the remainder of 4. Answer: E. Check Number theory chapter of Math Book for more: math-number-theory-88376.html Hope it helps. I just came to testing the cycle and felt a dead end. I am not getting why are we zeroing in on 3^3 for the first part of the addition in the question Intern Joined: 09 Apr 2013 Posts: 32 Location: United States Concentration: Social Entrepreneurship, Strategy GMAT 1: 750 Q50 V41 GPA: 3.55 Followers: 1 Kudos [?]: 38 [2] , given: 0 Re: Exponents [#permalink] Show Tags 06 Jun 2013, 15:10 2 This post received KUDOS 1 This post was BOOKMARKED pavan2185 wrote: Bunuel wrote: azule45 wrote: If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5? A 0 B 1 C 2 D 3 E 4 It should be $$3^{8n+3}+2$$. The units digit of 3 in positive integer power has cyclicity of 4 for the unis digit: 3^1 --> the units digit is 3; 3^2 --> the units digit is 9; 3^3 --> the units digit is 7; 3^4 --> the units digit is 1; 3^5 --> the units digit is 3 AGAIN; ... So, the units digit repeats the following pattern {3-9-7-1}-{3-9-7-1}-.... $$3^{8n+3}$$ will have the same units digit as $$3^3$$, which is 7 (remainder when 8n+3 divided by cyclicity 4 is 3). Thus the last digit of $$3^{8n+3}+2$$ will be $$7+2=9$$. Any positive integer with the unis digit of 9 divided by 5 gives the remainder of 4. Answer: E. Check Number theory chapter of Math Book for more: math-number-theory-88376.html Hope it helps. I just came to testing the cycle and felt a dead end. I am not getting why are we zeroing in on 3^3 for the first part of the addition in the question The reason we are focusing on $$3^3$$ is because the repeating cycle of $$3^{8n}$$ only matters because it is a sequence that loops in a pattern of 4 (which is consequently a divisor of 8). You said that you tested the cycle. So you should understand that, as Bunuel said, 3-9-7-1-3-9-7-1 is the cycle. No matter what multiple of 8 you use (8, 16, 24, 32, etc.), 1 will always be the units digit of the final value. If that is the case we can actually forget the 8n part of the equation, because you know that $$3^{8n}$$ will always have a units digit of 1. This means that you have to focus on the $$3^3$$ instead. You can look at this is two ways. Either go three more steps in the cycle and land on 7 (3-9-7), or because $$3^{8n+3} = 3^{8n}*3^3$$ = (number with the units digit of 1) x 27. The units digit will be 7 because 1 x 7 is 7. 7+2 = 9, 9/5 = 1 r4. Answer is E. You can also look at my post above for a method that doesn't involve all this sequencing stuff On another note, be careful not to make this the sequencing a rule for all questions that look like this because it definitely is not a rule. If you see $$3^{7m+3}$$, you can't just "cancel out" the 7m. If m=1 the cycle is 3-9-7-1-3-9-7, but if m=2 it continues the sequence and does not restart back at 3 but instead starts where it left off after 7 (1-3-9-7-1-3-9). The reason we focus on the $$3^3$$ in this equation is only because the final value of the "8n sequence" always ends with the same units digit. Kudos if that helps. Verbal Forum Moderator Joined: 10 Oct 2012 Posts: 630 Followers: 80 Kudos [?]: 1120 [2] , given: 136 Re: If n is a positive integer, what is the remainder when [#permalink] Show Tags 07 Jun 2013, 01:11 2 This post received KUDOS azule45 wrote: If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5? A. 0 B. 1 C. 2 D. 3 E. 4 We know that Remainder($$\frac{a*b}{c}$$) = Remainder($$\frac{a}{c}$$)*Remainder($$\frac{b}{c}$$) for integral values of a,b and c. The given expression : $$\frac{(3^{8n}*3^3 + 2)}{5}$$ = $$\frac{27*(3^4)^{2n}}{5}$$+$$\frac{2}{5}$$ = Remainder of $$\frac{(27)*(81)^{2n}}{5}$$ + Remainder of $$\frac{2}{5}$$ = Remainder of$$\frac{(2)*(1)^{2n}}{5}$$+Remainder of $$\frac{2}{5}$$= Remainder of $$\frac{4}{5}$$ = 4. _________________ Current Student Joined: 06 Sep 2013 Posts: 2035 Concentration: Finance GMAT 1: 770 Q0 V Followers: 62 Kudos [?]: 594 [0], given: 355 Re: If n is a positive integer, what is the remainder when [#permalink] Show Tags 03 Feb 2014, 09:32 azule45 wrote: If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5? A. 0 B. 1 C. 2 D. 3 E. 4 I followed a slightly longer approach Assume n=1 then 3^11 + 2 /5, what is the remainder? Well 3^11 is (5-2)^11 so we have -2^11/5 what is the remainder of this? Again -2*4^5 = -2*(5-1)^5, which gives a remainder of -1 multiplied by -2 gives a remainder of 2, and if we add the 2 that was missing for the first part of the question we get 4 as a remainder Hope its clear Cheers J Current Student Status: Everyone is a leader. Just stop listening to others. Joined: 22 Mar 2013 Posts: 992 Location: India GPA: 3.51 WE: Information Technology (Computer Software) Followers: 164 Kudos [?]: 1464 [0], given: 227 Re: If n is a positive integer, what is the remainder when [#permalink] Show Tags 09 Apr 2014, 07:03 $$\frac{(3^8)^n*3^3 + 2}{5}$$ $$\frac{81^n * 81^n * 27 + 2}{5}$$ Making use of binomial theorem: $$\frac{ (80+1)^n * (80+1)^n * 27 + 2 }{5}$$ $$\frac{ 1^n * 1^n * 2 + 2 }{5}$$ $$\frac{2+2}{5}$$ $$\frac{4}{5}$$ Thus 4 is remainder. _________________ Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos My Articles: 1. WOULD: when to use? | 2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction". Current Student Joined: 14 Jul 2013 Posts: 32 Followers: 0 Kudos [?]: 8 [0], given: 39 Re: If n is a positive integer, what is the remainder when [#permalink] Show Tags 11 Apr 2014, 05:06 [3^(8n+3) + 2]/5 --> Calc. Remainder => within 30 sec, check for n=1,=2,=3 numerator becomes (3^11+2),( 3^19+2),(3^27+2) 3 has cyclicity of 4 => Remainder from 11/4, 19/4, 27/4 = 3 from which last digit happens to be 7 therefore, we have drilled down the question to calculate remainder of (7+2)/5 i.e. 4 i.e. option E Intern Status: Preparing for GMAT Joined: 10 Dec 2013 Posts: 20 Location: India Concentration: Marketing, Leadership GMAT 1: 530 Q46 V18 WE: Other (Entertainment and Sports) Followers: 0 Kudos [?]: 7 [0], given: 61 Remainders and Divisibility [#permalink] Show Tags 28 Jul 2014, 17:12 If n is a positive integer, what is the remainder when 3^"8n+3" + 2 is divided by 5? A. 0 B. 1 C. 2 D. 3 E. 4 Manager Joined: 04 Sep 2012 Posts: 101 Location: Philippines Concentration: Marketing, Entrepreneurship Schools: Ross (Michigan) - Class of 2017 GMAT 1: 620 Q48 V27 GMAT 2: 660 Q47 V34 GMAT 3: 700 Q47 V38 GPA: 3.25 WE: Sales (Manufacturing) Followers: 1 Kudos [?]: 39 [0], given: 504 Re: Remainders and Divisibility [#permalink] Show Tags 28 Jul 2014, 17:22 1 This post was BOOKMARKED suhaschan wrote: If n is a positive integer, what is the remainder when 3^"8n+3" + 2 is divided by 5? A. 0 B. 1 C. 2 D. 3 E. 4 Get the cyclicity of 3 to get the units digit of the power of 3. Since n is a positive integer, Lets try picking a number n=1, 3^11 Cyclicity of 3 is 4 (3,9,27,81,243, repeat). Therefore n=1, 3^11 --> units digit is 7 7+2=9/5= Remainder of 4. Answer is E Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7125 Location: Pune, India Followers: 2137 Kudos [?]: 13689 [1] , given: 222 Re: Remainders and Divisibility [#permalink] Show Tags 28 Jul 2014, 20:54 1 This post received KUDOS Expert's post suhaschan wrote: If n is a positive integer, what is the remainder when 3^"8n+3" + 2 is divided by 5? A. 0 B. 1 C. 2 D. 3 E. 4 You should know the concept of cyclicity. Powers of 3 have a cyclicity of 4 i.e. 3^1 = 3 3^2 = 9 3^3 = 27 3^4 = 81 3^5 = 243 3^6 = 729 Look at the last digits of powers: 3, 9, 7, 1, 3, 9, 7, 1, 3, 9... So whenever we have a power of 3, we remove groups of 4 out of it i.e. divide it by 4 and whatever is the remainder tells us the last digit. So say we have 3^13, we divide 13 by 4 and are left with 1 remainder. So this is the first term of the next cycle and will end with 3. Say we have 3^39, we divide 39 by 4 and are left with 3 remainder. So this is the third term of the next cycle and will end with 7. If you have $$3^{8n+3}$$, we divide 8n+3 by 4 and are left with 3 remainder. So this is the third term of the next cycle and will end with 7. Last term of $$3^{8n+3}+2$$ will be 7+2 = 9. When you divide a number that ends with 9 by 5, you get 4 remainder. If you are not sure why this is so, check out the first part of the question discussed in this post: http://www.veritasprep.com/blog/2014/03 ... emainders/ Answer (E) Or since you know that cyclicity of 3 is 4, just put n = 0 to get 3^3 + 2 = 29. When you divide 29 by 5, you get remainder 4. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: If n is a positive integer, what is the remainder when [#permalink]

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29 Jul 2014, 00:36
suhaschan wrote:
If n is a positive integer, what is the remainder when 3^"8n+3" + 2 is divided by 5?

A. 0
B. 1
C. 2
D. 3
E. 4

Merging topics. Please refer to the discussion above.

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29 Jul 2014, 10:21
Thanks a lot for the patient and fantastic explanation.
Re: If n is a positive integer, what is the remainder when   [#permalink] 29 Jul 2014, 10:21

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