Aug 20 08:00 PM PDT  09:00 PM PDT EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth $100 with the 3 Month Pack ($299) Aug 20 09:00 PM PDT  10:00 PM PDT Take 20% off the plan of your choice, now through midnight on Tuesday, 8/20 Aug 22 09:00 PM PDT  10:00 PM PDT What you'll gain: Strategies and techniques for approaching featured GMAT topics, and much more. Thursday, August 22nd at 9 PM EDT Aug 24 07:00 AM PDT  09:00 AM PDT Learn reading strategies that can help even nonvoracious reader to master GMAT RC Aug 25 09:00 AM PDT  12:00 PM PDT Join a FREE 1day verbal workshop and learn how to ace the Verbal section with the best tips and strategies. Limited for the first 99 registrants. Register today! Aug 25 08:00 PM PDT  11:00 PM PDT Exclusive offer! Get 400+ Practice Questions, 25 Video lessons and 6+ Webinars for FREE.
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 57083

What is the remainder when (3^84)/26
[#permalink]
Show Tags
20 Oct 2015, 12:39
Question Stats:
70% (01:25) correct 30% (01:53) wrong based on 518 sessions
HideShow timer Statistics
What is the remainder when \(\frac{(3^{84})}{26}\) (A) 0 (B) 1 (C) 2 (D) 24 (E) 25
Official Answer and Stats are available only to registered users. Register/ Login.
_________________




CEO
Joined: 20 Mar 2014
Posts: 2620
Concentration: Finance, Strategy
GPA: 3.7
WE: Engineering (Aerospace and Defense)

Re: What is the remainder when (3^84)/26
[#permalink]
Show Tags
24 Oct 2015, 11:58
BrainLab wrote: Dear experts, do you think this is a valid method ? 3^84 has a unit digit 1, so the remainder is 1 Answer(B) Not necessarily. You did get the correct answer but with a 'lucky' approach. 3^4 also has a unit digit 1 but remainder of 3^4(=81) when you divide it by 26 is 3 and not 1. So for this and other similar questions try to find a number closest to either the divisor or a multiple of the divisor. You can clearly see that 3^3=27=26+1. Use this to your advantage as follows: \(3^{84} = (3^3)^{28} = (27)^{28} = (26+1)^{28}\) Applying the binomial theorem to expand \((a+b)^n = a^n*b^0+a^{n1}*b^1....+a^1*b^{n1}+a^0*b^n\), you get \((26+1)^{28} = 26^{28}+26^{27}*1^1...26^1*1^{27}+26^0*1^{28}\), so apart from the last term, all other terms have 26 and thus you will get a remainder of 0 with all but 1 terms in this expanded form. Thus the remainder will be = remainder of \(\frac{26^0*1^{28}}{26}\) = 1 B is the correct answer. Hope this helps.




Manager
Joined: 11 Sep 2013
Posts: 107

Re: What is the remainder when (3^84)/26
[#permalink]
Show Tags
20 Oct 2015, 13:05
Bunuel wrote: What is the remainder when \(\frac{(3^{84})}{26}\)
(A) 0 (B) 1 (C) 2 (D) 24 (E) 25 3^84/26 = 27^26/26 27 is divided by 26, the remainder is 1 => 27^26 is divided by 26, the remainder is still 1 Ans BPlease correct if i am wrong Thanks



Senior Manager
Joined: 10 Mar 2013
Posts: 485
Location: Germany
Concentration: Finance, Entrepreneurship
GPA: 3.88
WE: Information Technology (Consulting)

Re: What is the remainder when (3^84)/26
[#permalink]
Show Tags
24 Oct 2015, 11:43
Dear experts, do you think this is a valid method ? 3^84 has a unit digit 1, so the remainder is 1 Answer(B)
_________________
When you’re up, your friends know who you are. When you’re down, you know who your friends are.
Share some Kudos, if my posts help you. Thank you !
800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50 GMAT PREP 670 MGMAT CAT 630 KAPLAN CAT 660



Manager
Joined: 01 Jan 2015
Posts: 62

Re: What is the remainder when (3^84)/26
[#permalink]
Show Tags
24 Oct 2015, 13:32
BrainLab wrote: Dear experts, do you think this is a valid method ? 3^84 has a unit digit 1, so the remainder is 1 Answer(B) No, it is not a correct approach. You can use units digit to guide you in finding the remainder when dividing by 2, 5, or 10 for example. For example \(\frac{12}{5}\) and \(\frac{122222}{5}\) have the same remainder. The way I approach these types of remainder questions, is I first look at the divisor. If it is a small divisor that is neither 2,5, nor 10, I will choose between looking at the cyclicity of the base raised to a positive exponent divided by the divisor and binomial expansion. For example consider finding the remainder for \(\frac{2^{86}}{7}\). Since the divisor is not 2,5, or 10, looking at the units digit wouldn't work. If I were to use the cyclicity method, I would start by listing, \(\frac{2^{1}}{7}\)yields remainder of 2 \(\frac{2^{2}}{7}\)yields remainder of 4 \(\frac{2^{3}}{7}\)yields remainder of 1 \(\frac{2^{4}}{7}\)yields remainder of 2 There is a cyclicity of 3, since 86/3 yields remainder of 2 then the remainder of \(\frac{2^{86}}{7}\), will be the same remainder of \(\frac{2^{2}}{7}\),which as remainder of 4 Alternately, using binomial expansion method: \(\frac{2^{86}}{7}\) > \(\frac{(2^3)^{28}*2^2}{7}\) > \(\frac{(7+1)^{28}* 2^2}{7}\) Then I will look at the individual remainders of each of the 2 terms when dividing by 7. For example let's say you wanted to find the remainder of \(\frac{(18*23)}{7}\). Mod(\(\frac{18}{7}\)) is 4. Mod(\(\frac{23}{7}\)) is 2. Combine 4 and 2 by multiplication because the operation in question is multiplication and then find the remainder. So the remainder of \(\frac{(4*2)}{7}\) is 1 and this is the answer. This can be done with negative remainders too. For example let's say you wanted to find the remainder of \(\frac{(2318)}{7}\). Combine 2 and 4 by subtraction and then find the remainder. So the remainder of \(\frac{(24)}{7}\) is 5. When you have a negative remainder, find the remainder first as if the numerator was positive and then subtract from the divisor. If you had positive 2 on top, the remainder would have been 2, but you have negative 2, so subtract positive 2 from 7 and you get a remainder of 5. Back to the question. The remainder of \(\frac{(7+1)^{28}}{7}\) is 1; every term will be divisible by 7 except the last term,which is \(1^{28}.\) See Engr2012's above explanation regarding binomial expansion. The remainder of \(\frac{2^2}{7}\) is 4. Combine 1 and 4 by multiplication and then find the remainder. So the remainder of \(\frac{(1*4)}{7}\) is 4. Since the divisor in the original question is 26, the cyclicity would be too big. The cyclicity method wouldn't be a good idea. I would have done this question exactly as Engr2012 did.



Senior Manager
Joined: 10 Mar 2013
Posts: 485
Location: Germany
Concentration: Finance, Entrepreneurship
GPA: 3.88
WE: Information Technology (Consulting)

What is the remainder when (3^84)/26
[#permalink]
Show Tags
25 Oct 2015, 02:04
Thanks altogether, I've just tried this approach in 2 or 3 questions and did get a right answer, what made me think that this one is the fastest approach... but I saw my mistake now.
_________________
When you’re up, your friends know who you are. When you’re down, you know who your friends are.
Share some Kudos, if my posts help you. Thank you !
800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50 GMAT PREP 670 MGMAT CAT 630 KAPLAN CAT 660



Senior Manager
Joined: 10 Mar 2013
Posts: 485
Location: Germany
Concentration: Finance, Entrepreneurship
GPA: 3.88
WE: Information Technology (Consulting)

What is the remainder when (3^84)/26
[#permalink]
Show Tags
25 Oct 2015, 04:23
bhaskar438 wrote: BrainLab wrote: Dear experts, do you think this is a valid method ? 3^84 has a unit digit 1, so the remainder is 1 Answer(B) No, it is not a correct approach. You can use units digit to guide you in finding the remainder when dividing by 2, 5, or 10 for example. For example \(\frac{12}{5}\) and \(\frac{122222}{5}\) have the same remainder. The way I approach these types of remainder questions, is I first look at the divisor. If it is a small divisor that is neither 2,5, nor 10, I will choose between looking at the cyclicity of the base raised to a positive exponent divided by the divisor and binomial expansion. For example consider finding the remainder for \(\frac{2^{86}}{7}\). Since the divisor is not 2,5, or 10, looking at the units digit wouldn't work. If I were to use the cyclicity method, I would start by listing, \(\frac{2^{1}}{7}\)yields remainder of 2 \(\frac{2^{2}}{7}\)yields remainder of 4 \(\frac{2^{3}}{7}\)yields remainder of 1 \(\frac{2^{4}}{7}\)yields remainder of 2 There is a cyclicity of 3, since 86/3 yields remainder of 2 then the remainder of \(\frac{2^{86}}{7}\), will be the same remainder of \(\frac{2^{2}}{7}\),which as remainder of 4 Alternately, using binomial expansion method: \(\frac{2^{86}}{7}\) > \(\frac{(2^3)^{28}*2^2}{7}\) > \(\frac{(7+1)^{28}* 2^2}{7}\) Then I will look at the individual remainders of each of the 2 terms when dividing by 7. For example let's say you wanted to find the remainder of \(\frac{(18*23)}{7}\). Mod(\(\frac{18}{7}\)) is 4. Mod(\(\frac{23}{7}\)) is 2. Combine 4 and 2 by multiplication because the operation in question is multiplication and then find the remainder. So the remainder of \(\frac{(4*2)}{7}\) is 1 and this is the answer. This can be done with negative remainders too. For example let's say you wanted to find the remainder of \(\frac{(2318)}{7}\). Combine 2 and 4 by subtraction and then find the remainder. So the remainder of \(\frac{(24)}{7}\) is 5. When you have a negative remainder, find the remainder first as if the numerator was positive and then subtract from the divisor. If you had positive 2 on top, the remainder would have been 2, but you have negative 2, so subtract positive 2 from 7 and you get a remainder of 5. Back to the question. The remainder of \(\frac{(7+1)^{28}}{7}\) is 1; every term will be divisible by 7 except the last term,which is \(1^{28}.\) See Engr2012's above explanation regarding binomial expansion. The remainder of \(\frac{2^2}{7}\) is 4. Combine 1 and 4 by multiplication and then find the remainder. So the remainder of \(\frac{(1*4)}{7}\) is 4. Since the divisor in the original question is 26, the cyclicity would be too big. The cyclicity method wouldn't be a good idea. I would have done this question exactly as Engr2012 did. \(\frac{2^{86}}{7}\) isn't it the same using cyclicity method...? \(2^{86}/7\) has a unit's digit of 4 , so the remainder is 4 Engr2012, here we use cyclicity approach and it's valid. I just don't understand where is the limit where we can use or don't this approach ?? whatistheremainderof126493.html
_________________
When you’re up, your friends know who you are. When you’re down, you know who your friends are.
Share some Kudos, if my posts help you. Thank you !
800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50 GMAT PREP 670 MGMAT CAT 630 KAPLAN CAT 660



CEO
Joined: 20 Mar 2014
Posts: 2620
Concentration: Finance, Strategy
GPA: 3.7
WE: Engineering (Aerospace and Defense)

What is the remainder when (3^84)/26
[#permalink]
Show Tags
25 Oct 2015, 06:03
BrainLab wrote: bhaskar438 wrote:
No, it is not a correct approach. You can use units digit to guide you in finding the remainder when dividing by 2, 5, or 10 for example. For example \(\frac{12}{5}\) and \(\frac{122222}{5}\) have the same remainder. The way I approach these types of remainder questions, is I first look at the divisor. If it is a small divisor that is neither 2,5, nor 10, I will choose between looking at the cyclicity of the base raised to a positive exponent divided by the divisor and binomial expansion.
For example consider finding the remainder for \(\frac{2^{86}}{7}\). Since the divisor is not 2,5, or 10, looking at the units digit wouldn't work. If I were to use the cyclicity method, I would start by listing, \(\frac{2^{1}}{7}\)yields remainder of 2
\(\frac{2^{2}}{7}\)yields remainder of 4
\(\frac{2^{3}}{7}\)yields remainder of 1
\(\frac{2^{4}}{7}\)yields remainder of 2
There is a cyclicity of 3, since 86/3 yields remainder of 2 then the remainder of \(\frac{2^{86}}{7}\), will be the same remainder of \(\frac{2^{2}}{7}\),which as remainder of 4
Alternately, using binomial expansion method:
\(\frac{2^{86}}{7}\) > \(\frac{(2^3)^{28}*2^2}{7}\) > \(\frac{(7+1)^{28}* 2^2}{7}\)
Then I will look at the individual remainders of each of the 2 terms when dividing by 7. For example let's say you wanted to find the remainder of \(\frac{(18*23)}{7}\). Mod(\(\frac{18}{7}\)) is 4. Mod(\(\frac{23}{7}\)) is 2. Combine 4 and 2 by multiplication because the operation in question is multiplication and then find the remainder. So the remainder of \(\frac{(4*2)}{7}\) is 1 and this is the answer.
This can be done with negative remainders too. For example let's say you wanted to find the remainder of \(\frac{(2318)}{7}\). Combine 2 and 4 by subtraction and then find the remainder. So the remainder of \(\frac{(24)}{7}\) is 5. When you have a negative remainder, find the remainder first as if the numerator was positive and then subtract from the divisor. If you had positive 2 on top, the remainder would have been 2, but you have negative 2, so subtract positive 2 from 7 and you get a remainder of 5.
Back to the question. The remainder of \(\frac{(7+1)^{28}}{7}\) is 1; every term will be divisible by 7 except the last term,which is \(1^{28}.\) See Engr2012's above explanation regarding binomial expansion. The remainder of \(\frac{2^2}{7}\) is 4. Combine 1 and 4 by multiplication and then find the remainder. So the remainder of \(\frac{(1*4)}{7}\) is 4.
Since the divisor in the original question is 26, the cyclicity would be too big. The cyclicity method wouldn't be a good idea. I would have done this question exactly as Engr2012 did. \(\frac{2^{86}}{7}\) isn't it the same using cyclicity method...? \(2^{86}/7\) has a unit's digit of 4 , so the remainder is 4 Engr2012, here we use cyclicity approach and it's valid. I just don't understand where is the limit where we can use or don't this approach ?? whatistheremainderof126493.htmlFirst off, 1 method "working" for a particular question does not mean that it is the most efficient / repeatable method. \(2^{86}/7\) luckily gave you the correct answer but you want to develop and understand a method that can be applied to all such questions. Cyclicity is not such a method. Cyclicity is to be used for finding out what will be the unit's digit but for questions similar to \(2^{86}/7\), I will use the binomial theorem route. \(2^{86}/7\) = \(2^2*(2^6)^{14}\) =\(4*(64)^{14}\) = \(4*(63+1)^{14}\) Again, all terms of (63+1)^14 will have 63 (a multiple of 7) except the last one (=1^14), giving you \(4*1^{14}\) = 4 (as the last term). Thus the remainder when you divide 4/7=4. Idea behind 2^6 is to find the closest multiple of 7 (such that the number = multiple of 7 + 1), alternately you could have used 2^3 as well as it will give you 8 = 71. Concept remains the same. Similarly, to generalize this concept, 2^n needs to be expressed in the form of (divisor+1)^n in order to use the binomial theorem. Hope this helps.



Senior Manager
Joined: 10 Mar 2013
Posts: 485
Location: Germany
Concentration: Finance, Entrepreneurship
GPA: 3.88
WE: Information Technology (Consulting)

Re: What is the remainder when (3^84)/26
[#permalink]
Show Tags
25 Oct 2015, 06:14
Engr2012 wrote: BrainLab wrote: bhaskar438 wrote:
No, it is not a correct approach. You can use units digit to guide you in finding the remainder when dividing by 2, 5, or 10 for example. For example \(\frac{12}{5}\) and \(\frac{122222}{5}\) have the same remainder. The way I approach these types of remainder questions, is I first look at the divisor. If it is a small divisor that is neither 2,5, nor 10, I will choose between looking at the cyclicity of the base raised to a positive exponent divided by the divisor and binomial expansion.
For example consider finding the remainder for \(\frac{2^{86}}{7}\). Since the divisor is not 2,5, or 10, looking at the units digit wouldn't work. If I were to use the cyclicity method, I would start by listing, \(\frac{2^{1}}{7}\)yields remainder of 2
\(\frac{2^{2}}{7}\)yields remainder of 4
\(\frac{2^{3}}{7}\)yields remainder of 1
\(\frac{2^{4}}{7}\)yields remainder of 2
There is a cyclicity of 3, since 86/3 yields remainder of 2 then the remainder of \(\frac{2^{86}}{7}\), will be the same remainder of \(\frac{2^{2}}{7}\),which as remainder of 4
Alternately, using binomial expansion method:
\(\frac{2^{86}}{7}\) > \(\frac{(2^3)^{28}*2^2}{7}\) > \(\frac{(7+1)^{28}* 2^2}{7}\)
Then I will look at the individual remainders of each of the 2 terms when dividing by 7. For example let's say you wanted to find the remainder of \(\frac{(18*23)}{7}\). Mod(\(\frac{18}{7}\)) is 4. Mod(\(\frac{23}{7}\)) is 2. Combine 4 and 2 by multiplication because the operation in question is multiplication and then find the remainder. So the remainder of \(\frac{(4*2)}{7}\) is 1 and this is the answer.
This can be done with negative remainders too. For example let's say you wanted to find the remainder of \(\frac{(2318)}{7}\). Combine 2 and 4 by subtraction and then find the remainder. So the remainder of \(\frac{(24)}{7}\) is 5. When you have a negative remainder, find the remainder first as if the numerator was positive and then subtract from the divisor. If you had positive 2 on top, the remainder would have been 2, but you have negative 2, so subtract positive 2 from 7 and you get a remainder of 5.
Back to the question. The remainder of \(\frac{(7+1)^{28}}{7}\) is 1; every term will be divisible by 7 except the last term,which is \(1^{28}.\) See Engr2012's above explanation regarding binomial expansion. The remainder of \(\frac{2^2}{7}\) is 4. Combine 1 and 4 by multiplication and then find the remainder. So the remainder of \(\frac{(1*4)}{7}\) is 4.
Since the divisor in the original question is 26, the cyclicity would be too big. The cyclicity method wouldn't be a good idea. I would have done this question exactly as Engr2012 did. \(\frac{2^{86}}{7}\) isn't it the same using cyclicity method...? \(2^{86}/7\) has a unit's digit of 4 , so the remainder is 4 Engr2012, here we use cyclicity approach and it's valid. I just don't understand where is the limit where we can use or don't this approach ?? whatistheremainderof126493.htmlFirst off, 1 method "working" for a particular question does not mean that it is the most efficient / repeatable method. \(2^{86}/7\) luckily gave you the correct answer but you want to develop and understand a method that can be applied to all such questions. Cyclicity is not such a method. Cyclicity is to be used for finding out what will be the unit's digit but for questions similar to \(2^{86}/7\), I will use the binomial theorem route. \(2^{86}/7\) = \(2^2*(2^6)^{14}\) =\(4*(64)^{14}\) = \(4*(63+1)^{14}\) Again, all terms of (63+1)^14 will have 63 (a multiple of 7) except the last one (=1^14), giving you \(4*1^{14}\) = 4 (as the last term). Thus the remainder when you divide 4/7=4. Idea behind 2^6 is to find the closest multiple of 7 (such that the number = multiple of 7 + 1), alternately you could have used 2^3 as well as it will give you 8 = 71. Concept remains the same. Similarly, to generalize this concept, 2^n needs to be expressed in the form of (divisor+1)^n in order to use the binomial theorem. Hope this helps. Hi ENGR2012, first of all , thanks for your time. After investigating about this topic, I must say that cyclicity is a valid approach (as bhaskar438 have already stated) only if we have 2,5 or 10 as a divisor. But I'll better use binomial theorem to be 100% sure.
_________________
When you’re up, your friends know who you are. When you’re down, you know who your friends are.
Share some Kudos, if my posts help you. Thank you !
800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50 GMAT PREP 670 MGMAT CAT 630 KAPLAN CAT 660



CEO
Joined: 20 Mar 2014
Posts: 2620
Concentration: Finance, Strategy
GPA: 3.7
WE: Engineering (Aerospace and Defense)

What is the remainder when (3^84)/26
[#permalink]
Show Tags
25 Oct 2015, 06:17
BrainLab wrote: Hi ENGR2012, first of all , thanks for your time. After investigating about this topic, I must say that cyclicity is a valid approach (as bhaskar438 have already stated) only if we have 2,5 or 10 as a divisor. But I'll better use binomial theorem to be 100% sure. GMAT is all about picking the best method in terms of execution and time you spend on a particular question. If you are comfortable to add the extra bit of learning that 2,5,10 maintain cyclicity then by all means do it but make sure to not experiment in the GMAT. I never said that cyclicity is wrong. For me, binomial theorem application is more straightforward. Pick your method and stick to it. Pattern recognition and application of a fixed method of attack are of paramount importance in GMAT.



Manager
Joined: 01 Jan 2015
Posts: 62

Re: What is the remainder when (3^84)/26
[#permalink]
Show Tags
25 Oct 2015, 07:33
BrainLab wrote: bhaskar438 wrote: BrainLab wrote: Dear experts, do you think this is a valid method ? 3^84 has a unit digit 1, so the remainder is 1 Answer(B) No, it is not a correct approach. You can use units digit to guide you in finding the remainder when dividing by 2, 5, or 10 for example. For example \(\frac{12}{5}\) and \(\frac{122222}{5}\) have the same remainder. The way I approach these types of remainder questions, is I first look at the divisor. If it is a small divisor that is neither 2,5, nor 10, I will choose between looking at the cyclicity of the base raised to a positive exponent divided by the divisor and binomial expansion. For example consider finding the remainder for \(\frac{2^{86}}{7}\). Since the divisor is not 2,5, or 10, looking at the units digit wouldn't work. If I were to use the cyclicity method, I would start by listing, \(\frac{2^{1}}{7}\)yields remainder of 2 \(\frac{2^{2}}{7}\)yields remainder of 4 \(\frac{2^{3}}{7}\)yields remainder of 1 \(\frac{2^{4}}{7}\)yields remainder of 2 There is a cyclicity of 3, since 86/3 yields remainder of 2 then the remainder of \(\frac{2^{86}}{7}\), will be the same remainder of \(\frac{2^{2}}{7}\),which as remainder of 4 Alternately, using binomial expansion method: \(\frac{2^{86}}{7}\) > \(\frac{(2^3)^{28}*2^2}{7}\) > \(\frac{(7+1)^{28}* 2^2}{7}\) Then I will look at the individual remainders of each of the 2 terms when dividing by 7. For example let's say you wanted to find the remainder of \(\frac{(18*23)}{7}\). Mod(\(\frac{18}{7}\)) is 4. Mod(\(\frac{23}{7}\)) is 2. Combine 4 and 2 by multiplication because the operation in question is multiplication and then find the remainder. So the remainder of \(\frac{(4*2)}{7}\) is 1 and this is the answer. This can be done with negative remainders too. For example let's say you wanted to find the remainder of \(\frac{(2318)}{7}\). Combine 2 and 4 by subtraction and then find the remainder. So the remainder of \(\frac{(24)}{7}\) is 5. When you have a negative remainder, find the remainder first as if the numerator was positive and then subtract from the divisor. If you had positive 2 on top, the remainder would have been 2, but you have negative 2, so subtract positive 2 from 7 and you get a remainder of 5. Back to the question. The remainder of \(\frac{(7+1)^{28}}{7}\) is 1; every term will be divisible by 7 except the last term,which is \(1^{28}.\) See Engr2012's above explanation regarding binomial expansion. The remainder of \(\frac{2^2}{7}\) is 4. Combine 1 and 4 by multiplication and then find the remainder. So the remainder of \(\frac{(1*4)}{7}\) is 4. Since the divisor in the original question is 26, the cyclicity would be too big. The cyclicity method wouldn't be a good idea. I would have done this question exactly as Engr2012 did. \(\frac{2^{86}}{7}\) isn't it the same using cyclicity method...? \(2^{86}/7\) has a unit's digit of 4 , so the remainder is 4 Engr2012, here we use cyclicity approach and it's valid. I just don't understand where is the limit where we can use or don't this approach ?? whatistheremainderof126493.htmlHi Brainlab, I was actually talking about 3 separate methods: units digit method, cycle method, and binomial expansion method. A specific units digit method for divisors 2,5,10. In the cyclicity method you are only looking at the cycle of the units digit to see when it stops repeating. You are not looking solely at the units digit to find the remainder as you would for divisors 2,5,10. In the previous question \(\frac{2^{86}}{7}\) was similar to finding the remainder for \(\frac{2^{2}}{7}\) because I found out the remainder should be the second item out of the three items in the cycle. Only use the cyclicity method when the divisor is small.



Manager
Joined: 01 Jan 2015
Posts: 62

Re: What is the remainder when (3^84)/26
[#permalink]
Show Tags
25 Oct 2015, 07:42
Engr2012 wrote: BrainLab wrote: Hi ENGR2012, first of all , thanks for your time. After investigating about this topic, I must say that cyclicity is a valid approach (as bhaskar438 have already stated) only if we have 2,5 or 10 as a divisor. But I'll better use binomial theorem to be 100% sure. GMAT is all about picking the best method in terms of execution and time you spend on a particular question. If you are comfortable to add the extra bit of learning that 2,5,10 maintain cyclicity then by all means do it but make sure to not experiment in the GMAT. I never said that cyclicity is wrong. For me, binomial theorem application is more straightforward. Pick your method and stick to it. Pattern recognition and application of a fixed method of attack are of paramount importance in GMAT. Sometimes the most efficient method is using binomial expansion first and then the cyclicity method as Bunuel as done in this question: whatistheremainderwhen182210isdividedby99724.html#p768816I can't think of a better method to find the remainder for \(\frac{18^{220}}{7}\) than the method he did. Also the cyclicity method is different from units digit method (which is for divisors 2,5, or 10)



CEO
Joined: 20 Mar 2014
Posts: 2620
Concentration: Finance, Strategy
GPA: 3.7
WE: Engineering (Aerospace and Defense)

What is the remainder when (3^84)/26
[#permalink]
Show Tags
25 Oct 2015, 07:55
bhaskar438 wrote: Engr2012 wrote: BrainLab wrote: Hi ENGR2012, first of all , thanks for your time. After investigating about this topic, I must say that cyclicity is a valid approach (as bhaskar438 have already stated) only if we have 2,5 or 10 as a divisor. But I'll better use binomial theorem to be 100% sure. GMAT is all about picking the best method in terms of execution and time you spend on a particular question. If you are comfortable to add the extra bit of learning that 2,5,10 maintain cyclicity then by all means do it but make sure to not experiment in the GMAT. I never said that cyclicity is wrong. For me, binomial theorem application is more straightforward. Pick your method and stick to it. Pattern recognition and application of a fixed method of attack are of paramount importance in GMAT. Sometimes the most efficient method is using binomial expansion first and then the cyclicity method as Bunuel as done in this question: whatistheremainderwhen182210isdividedby99724.html#p768816I can't think of a better method to find the remainder for \(\frac{18^{220}}{7}\) than the method he did. Also the cyclicity method is different from units digit method (which is for divisors 2,5, or 10) I dont think such a question will be asked in GMAT (not right now, even Bunuel mentions this in the first line of his solution) as I have not seen a single official/GMAT question that can only be solved by binomial theorem application. Almost always there were other ways to solve such a question. It is not about how many "difficult" questions you solve, there can be hundreds of them but the more important part of GMAT quant is to understand the tactics to solve 'problem groups'.



Manager
Joined: 01 Jan 2015
Posts: 62

Re: What is the remainder when (3^84)/26
[#permalink]
Show Tags
25 Oct 2015, 09:04
Engr2012 wrote: I dont think such a question will be asked in GMAT (not right now, even Bunuel mentions this in the first line of his solution) as I have not seen a single official/GMAT question that can only be solved by binomial theorem application. Almost always there were other ways to solve such a question. It is not about how many "difficult" questions you solve, there can be hundreds of them but the more important part of GMAT quant is to understand the tactics to solve 'problem groups'. I completely agree. I just like having many different methods in my arsenal.



Intern
Joined: 24 Nov 2015
Posts: 16
Location: India

Re: What is the remainder when (3^84)/26
[#permalink]
Show Tags
26 Nov 2015, 04:11
Such Questions Usually have a sort of Repetition in them :
3^1 = 3/26 gives remainder 3 3^2 = 9/26 gives remainder 9 3^3 = 27/26 gives remainder 1 3^4 = 81/26 again gives remainder 3 3^5 = 243/26 again gives remainder 9 and so on.
Its following a repetion of 3...9...1 or we can say they are repeating in multiples of 3. So, we can simply divide 84 by 3 which is completely divisible.
=> As per the repetition the remainder should be 1.



Manager
Joined: 08 Oct 2016
Posts: 201
Location: United States
Concentration: General Management, Finance
GPA: 2.9
WE: Engineering (Telecommunications)

Re: What is the remainder when (3^84)/26
[#permalink]
Show Tags
14 Apr 2017, 08:34
Bunuel wrote: What is the remainder when \(\frac{(3^{84})}{26}\)
(A) 0 (B) 1 (C) 2 (D) 24 (E) 25 BunuelAs per cyclicity it is 4m+4 mean last 2 digits are 81 so 81/28=78 Answer should be 3 isnot it? like this question https://gmatclub.com/forum/whatisthe ... 34109.html
_________________
Got Q42,V17 Target#01 Q45,V20April End



Math Expert
Joined: 02 Sep 2009
Posts: 57083

Re: What is the remainder when (3^84)/26
[#permalink]
Show Tags
14 Apr 2017, 08:37
ehsan090 wrote: Bunuel wrote: What is the remainder when \(\frac{(3^{84})}{26}\)
(A) 0 (B) 1 (C) 2 (D) 24 (E) 25 BunuelAs per cyclicity it is 4m+4 mean last 2 digits are 81 so 81/28=78 Answer should be 3 isnot it? like this question https://gmatclub.com/forum/whatisthe ... 34109.htmlNo, it's 1: https://www.wolframalpha.com/input/?i=t ... of+(3%5E84)%2F7 Please show your work.
_________________



Current Student
Joined: 19 Aug 2016
Posts: 148
Location: India
GPA: 3.82

Re: What is the remainder when (3^84)/26
[#permalink]
Show Tags
28 Jun 2017, 09:53
Bunuel wrote: What is the remainder when \(\frac{(3^{84})}{26}\)
(A) 0 (B) 1 (C) 2 (D) 24 (E) 25 Hi Bunuel, Will it be possible for you to provide an approach to this type of questions? or could you guide me to a post where you have already posted an explanation to such questions? I read the entire thread and I was unable to understand the approaches and frankly it seemed like a really time consuming approaches. Kindly help, thanks in advance!
_________________
Consider giving me Kudos if you find my posts useful, challenging and helpful!



Math Expert
Joined: 02 Sep 2009
Posts: 57083

What is the remainder when (3^84)/26
[#permalink]
Show Tags
28 Jun 2017, 10:21
ashikaverma13 wrote: Bunuel wrote: What is the remainder when \(\frac{(3^{84})}{26}\)
(A) 0 (B) 1 (C) 2 (D) 24 (E) 25 Hi Bunuel, Will it be possible for you to provide an approach to this type of questions? or could you guide me to a post where you have already posted an explanation to such questions? I read the entire thread and I was unable to understand the approaches and frankly it seemed like a really time consuming approaches. Kindly help, thanks in advance! Check below: REMAINDERS ON THE GMATDivisibility and Remainders on the GMATDivisibility TipsTheory on remainders problemsTips on remaindersUnits digits, exponents, remainders problemsDS remainders problemsPS remainders problemsDS divisibility problemsPS divisibility problemsHope it helps.
_________________



Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 4571
Location: India
GPA: 3.5
WE: Business Development (Commercial Banking)

What is the remainder when (3^84)/26
[#permalink]
Show Tags
28 Jun 2017, 11:17
Bunuel wrote: What is the remainder when \(\frac{(3^{84})}{26}\)
(A) 0 (B) 1 (C) 2 (D) 24 (E) 25 \(\frac{(3^{84})}{26}\) = \(\frac{3^{3*28}}{26}\) = \(\frac{27^{28}}{26}\) = \(\frac{(26 + 1)^{28}}{26}\) = \(\frac{(26^{28} + 1^{28})}{26}\) \frac{26^{Any \ Number }}{26} , will have remainder 0 \frac{1^{Any \ Number }}{26} , will have remainder 1 So, The answer will be (B) 1
_________________




What is the remainder when (3^84)/26
[#permalink]
28 Jun 2017, 11:17



Go to page
1 2
Next
[ 22 posts ]



