Author 
Message 
Senior Manager
Joined: 02 Dec 2007
Posts: 449

remainder [#permalink]
Show Tags
23 Jul 2008, 05:34
This topic is locked. If you want to discuss this question please repost it in the respective forum.
Guys can anyone explain how to approach such questions 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
3^3 +2 / 5 = 4 . Is this approach correct?



Current Student
Joined: 12 Jun 2008
Posts: 287
Schools: INSEAD Class of July '10

Re: remainder [#permalink]
Show Tags
23 Jul 2008, 05:58
Nihit wrote: Guys can anyone explain how to approach such questions 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
3^3 +2 / 5 = 4 . Is this approach correct? 4 is indeed the correct answer. The remainder of 3^1 when divided by 5 is 3 The remainder of 3^2 when divided by 5 is 4 The remainder of 3^3 when divided by 5 is 2 The remainder of 3^4 when divided by 5 is 1 When you reach "1" it means that the period you described is going to repeat over and over. Therefore: The remainder of 3^(4k+1) when divided by 5 is 3, with k any nonnegative integer The remainder of 3^(4k+2) when divided by 5 is 4, with k any nonnegative integer The remainder of 3^(4k+3) when divided by 5 is 2, with k any nonnegative integer The remainder of 3^(4k+4) when divided by 5 is 1, with k any nonnegative integer Here: 3^(8n+3) = 3^(4*(2n)+3) so the remainder when divided by 5 is 2 Add 2 to this number and the remainder when divided by 5 becomes 4 ==> Answer is (E)



Senior Manager
Joined: 06 Apr 2008
Posts: 435

Re: remainder [#permalink]
Show Tags
23 Jul 2008, 07:33
Nihit wrote: Guys can anyone explain how to approach such questions 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
3^3 +2 / 5 = 4 . Is this approach correct? The way I would tackle this Q is following If you see 3^n then unit digit gets repeated after every four numbers 3, 9 , 27, 81, 243 Therefore 3 ^ (8n + 3) means last digit is 7 which when added by 2 becomes 9. And remainder of any number with unit digit 9 is 4



Retired Moderator
Joined: 18 Jul 2008
Posts: 970

Re: remainder [#permalink]
Show Tags
23 Jul 2008, 17:12
Therefore 3 ^ (8n + 3) means last digit is 7 which when added by 2 becomes 9I do not understand why this is done (the purpose)  can you assist? thx nmohindru wrote: Nihit wrote: Guys can anyone explain how to approach such questions 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
3^3 +2 / 5 = 4 . Is this approach correct? The way I would tackle this Q is following If you see 3^n then unit digit gets repeated after every four numbers 3, 9 , 27, 81, 243 Therefore 3 ^ (8n + 3) means last digit is 7 which when added by 2 becomes 9. And remainder of any number with unit digit 9 is 4



Intern
Joined: 20 Jul 2008
Posts: 24

Re: remainder [#permalink]
Show Tags
23 Jul 2008, 20:43
nmohindru wrote: Nihit wrote: Guys can anyone explain how to approach such questions 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
3^3 +2 / 5 = 4 . Is this approach correct? The way I would tackle this Q is following If you see 3^n then unit digit gets repeated after every four numbers 3, 9 , 27, 81, 243 Therefore 3 ^ (8n + 3) means last digit is 7 which when added by 2 becomes 9. And remainder of any number with unit digit 9 is 4 Can you please explain the second part?



Manager
Joined: 27 May 2008
Posts: 141

Re: remainder [#permalink]
Show Tags
23 Jul 2008, 20:47
Its a typical Remainder theorem problem.
Here's my approach Remainder 3^1/5 = 3 3^2/5 = 4 3^3/5 = 2 3^4/5 = 1
Now every power of the order of 3^4n will give a remainder of 1.
3^(8n+3) = 3^8n. 3^3
Now as per the remainder theorem:
the remainders could of 3^8n and 3^3 can be multiplied.
3^8n will give a remainder of 1, 3^3 will give a remainder of 2.
1*2+2/5 will give a remainder of 4.



Senior Manager
Joined: 19 Mar 2008
Posts: 351

Re: remainder [#permalink]
Show Tags
24 Jul 2008, 00:20
Consider the unit digit of the index of 3, 3^1 is 3 3^2 is 9 3^3 is 7 3^4 is 1
and then repeating 3, 9, 7, 1
So the unit digit of 3^(8n+3) is 7
So the unit digit of 3^(8n+3) +2 is 9
Which gives the reminder of 4 when divided by 5
(E)



Current Student
Joined: 12 Jun 2008
Posts: 287
Schools: INSEAD Class of July '10

Re: remainder [#permalink]
Show Tags
24 Jul 2008, 00:24
The "unit digit repeats" method only works because the exercise asks to divide by 5. If it were by 4, 6, 7, etc... it wouldn't work.
That's why I think the "reminder repeats" method is better since it is a general method.



Manager
Joined: 14 Jun 2008
Posts: 161

Re: remainder [#permalink]
Show Tags
24 Jul 2008, 03:47
3^(8n+3) + 2
= 3 ^(4N) * 3 * (3) + 2 = (3 ^ 4)^N * 9 + 2 = (81)^N * 3 * 9 + 2
81 ^ N will always have a 1 in the units place. so when multiplied by 3, will have 3 in the units place which so when multiplied by 9, will have 7 in the units place
adding two will give a 9 in the units place.
any integer with a 9 in the units place, when divided by 5 will give a remainder of 4
Hence D
OA?



Retired Moderator
Joined: 18 Jul 2008
Posts: 970

Re: remainder [#permalink]
Show Tags
24 Jul 2008, 06:11
rahulgoyal1986  1*2+2/5 will give a remainder of 4.But: 2/5 gives remainder of 3. 3+2 = 5 Did I miss something? rahulgoyal1986 wrote: Its a typical Remainder theorem problem.
Here's my approach Remainder 3^1/5 = 3 3^2/5 = 4 3^3/5 = 2 3^4/5 = 1
Now every power of the order of 3^4n will give a remainder of 1.
3^(8n+3) = 3^8n. 3^3
Now as per the remainder theorem:
the remainders could of 3^8n and 3^3 can be multiplied.
3^8n will give a remainder of 1, 3^3 will give a remainder of 2.
1*2+2/5 will give a remainder of 4.



Current Student
Joined: 12 Jun 2008
Posts: 287
Schools: INSEAD Class of July '10

Re: remainder [#permalink]
Show Tags
24 Jul 2008, 06:58
bigfernhead wrote: rahulgoyal1986 
1*2+2/5 will give a remainder of 4.
But:
2/5 gives remainder of 3.
3+2 = 5
Did I miss something? This is just that 2/5 gives a remainder of 2 and not 3 2 = 0*5 + 2



Manager
Joined: 27 May 2008
Posts: 141

Re: remainder [#permalink]
Show Tags
24 Jul 2008, 07:34
judokan wrote: Consider the unit digit of the index of 3, 3^1 is 3 3^2 is 9 3^3 is 7 3^4 is 1
and then repeating 3, 9, 7, 1
So the unit digit of 3^(8n+3) is 7
So the unit digit of 3^(8n+3) +2 is 9
Which gives the reminder of 4 when divided by 5
(E) Wrong approach: Here's the way to proove it 4^24/7 find the remainder? Now as per your approach the answer should be 6. But the actual answer is 1. Say 4^1/7 = 4 4^2 = 16 4^3= 64 Thus 4^24 can be written as (4^3).(4^3).(4^3)...8 times, Every power of 4^3/7 will give a remainder of 1. 1.1.1.1...8 times divided by 7 will give a remainder of 1.



Manager
Joined: 27 May 2008
Posts: 141

Re: remainder [#permalink]
Show Tags
24 Jul 2008, 07:36
Oski wrote: bigfernhead wrote: rahulgoyal1986 
1*2+2/5 will give a remainder of 4.
But:
2/5 gives remainder of 3.
3+2 = 5
Did I miss something? This is just that 2/5 gives a remainder of 2 and not 3 2 = 0*5 + 2 Oski1*2+2/5 means 4 on Nr and 5 on Dr. thus a remainder of 4. If still not clear feel free to quote again



Manager
Joined: 27 May 2008
Posts: 141

Re: remainder [#permalink]
Show Tags
24 Jul 2008, 07:37
sset009 wrote: 3^(8n+3) + 2
= 3 ^(4N) * 3 * (3) + 2 = (3 ^ 4)^N * 9 + 2 = (81)^N * 3 * 9 + 2
81 ^ N will always have a 1 in the units place. so when multiplied by 3, will have 3 in the units place which so when multiplied by 9, will have 7 in the units place
adding two will give a 9 in the units place.
any integer with a 9 in the units place, when divided by 5 will give a remainder of 4
Hence D
OA? Wrong approach: Here's the way to proove it 4^24/7 find the remainder? Now as per your approach the answer should be 6. But the actual answer is 1. Say 4^1/7 = 4 4^2 = 16 4^3= 64 Thus 4^24 can be written as (4^3).(4^3).(4^3)...8 times, Every power of 4^3/7 will give a remainder of 1. 1.1.1.1...8 times divided by 7 will give a remainder of 1.



Current Student
Joined: 12 Jun 2008
Posts: 287
Schools: INSEAD Class of July '10

Re: remainder [#permalink]
Show Tags
24 Jul 2008, 08:21
rahulgoyal1986 wrote: Oski1*2+2/5 means 4 on Nr and 5 on Dr. thus a remainder of 4. If still not clear feel free to quote again I was just quoting the "2/5 gives remainder of 3" 2/5 gives a remainder of 2, and not 3 (that's what I was saying above )
Last edited by Oski on 24 Jul 2008, 08:52, edited 1 time in total.



Retired Moderator
Joined: 18 Jul 2008
Posts: 970

Re: remainder [#permalink]
Show Tags
24 Jul 2008, 08:35
Thanks gang I see my mistake.



Manager
Joined: 14 Jun 2008
Posts: 161

Re: remainder [#permalink]
Show Tags
24 Jul 2008, 12:13
rahulgoyal1986 wrote: Wrong approach: Here's the way to proove it 4^24/7 find the remainder?
Now as per your approach the answer should be 6. But the actual answer is 1. Say 4^1/7 = 4 4^2 = 16 4^3= 64
Thus 4^24 can be written as (4^3).(4^3).(4^3)...8 times, Every power of 4^3/7 will give a remainder of 1. 1.1.1.1...8 times divided by 7 will give a remainder of 1.
thanx, i think i see the mistake could you please state the remainder theorem? cheers



Senior Manager
Joined: 06 Apr 2008
Posts: 435

Re: remainder [#permalink]
Show Tags
25 Jul 2008, 00:06
rahulgoyal1986 wrote: nmohindru wrote: Nihit wrote: Guys can anyone explain how to approach such questions 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
3^3 +2 / 5 = 4 . Is this approach correct? The way I would tackle this Q is following If you see 3^n then unit digit gets repeated after every four numbers 3, 9 , 27, 81, 243 Wrong approach: Here's the way to proove it 4^24/7 find the remainder? Now as per your approach the answer should be 6. But the actual answer is 1. Say 4^1/7 = 4 4^2 = 16 4^3= 64 Thus 4^24 can be written as (4^3).(4^3).(4^3)...8 times, Every power of 4^3/7 will give a remainder of 1. 1.1.1.1...8 times divided by 7 will give a remainder of 1. Therefore 3 ^ (8n + 3) means last digit is 7 which when added by 2 becomes 9. And remainder of any number with unit digit 9 is 4 Well hold on ... where did I say that 4^n gets repeated ? Each specific number has its own properties e.g. if number has to be divisible by 3 then total of digits should be divisible by 3. This does not mean that if total of digits is divisible by 4 then it is divisible by 4 as well. 4 has its own properties. I hope you get what I mean



Manager
Joined: 27 May 2008
Posts: 141

Re: remainder [#permalink]
Show Tags
25 Jul 2008, 00:19
@ nmohindru
Frd
I cud not understand your approach. Kindly solve 4^24/7 for the remainder from your approach and post. Probably you have an easier and shorter way.



Senior Manager
Joined: 06 Apr 2008
Posts: 435

Re: remainder [#permalink]
Show Tags
25 Jul 2008, 00:34
rahulgoyal1986 wrote: @ nmohindru
Frd
I cud not understand your approach. Kindly solve 4^24/7 for the remainder from your approach and post. Probably you have an easier and shorter way. Rahul you are trying to compare apples with oranges When divisor is 5 no matter what is tens digit the units digit always determine the remainder. In case of 7 this does not hold true so approach will be different. Each number's specific properties regarding their squares, remainders etc are applied after seeing a particular Q and is only used to speed up that particular Q and should not be used to solve every Q of that nature. Take a look at example below If Q is whether 249678 is divisible by 11 then I would find sum of each even digit and compare with sum of each odd digit. If sum is equal that means number is divisible by 11. Now if Q is whether same number is divisible by 9 then I will sum each digit and see whether sum is divisible by 9 or not. So different approaches allow me to get the solution faster. A consistent approach will be to divide the number and see which will give you correct answers for all the similar Qs.







Go to page
1 2
Next
[ 23 posts ]



