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remainder [#permalink] New post 23 Jul 2008, 04:34
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?
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Re: remainder [#permalink] New post 23 Jul 2008, 04:58
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 non-negative integer
The remainder of 3^(4k+2) when divided by 5 is 4, with k any non-negative integer
The remainder of 3^(4k+3) when divided by 5 is 2, with k any non-negative integer
The remainder of 3^(4k+4) when divided by 5 is 1, with k any non-negative 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)
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Re: remainder [#permalink] New post 23 Jul 2008, 06:33
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
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Re: remainder [#permalink] New post 23 Jul 2008, 16:12
Therefore 3 ^ (8n + 3) means last digit is 7 which when added by 2 becomes 9

I 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
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Re: remainder [#permalink] New post 23 Jul 2008, 19:43
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?
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Re: remainder [#permalink] New post 23 Jul 2008, 19:47
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.
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Re: remainder [#permalink] New post 23 Jul 2008, 23:20
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)
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Re: remainder [#permalink] New post 23 Jul 2008, 23:24
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.
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Re: remainder [#permalink] New post 24 Jul 2008, 02:47
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?
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Re: remainder [#permalink] New post 24 Jul 2008, 05:11
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.
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Re: remainder [#permalink] New post 24 Jul 2008, 05:58
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
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Re: remainder [#permalink] New post 24 Jul 2008, 06:34
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.
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Re: remainder [#permalink] New post 24 Jul 2008, 06:36
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


Oski

1*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 :-D
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Re: remainder [#permalink] New post 24 Jul 2008, 06:37
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.
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Re: remainder [#permalink] New post 24 Jul 2008, 07:21
rahulgoyal1986 wrote:
Oski

1*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 :-D

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, 07:52, edited 1 time in total.
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Re: remainder [#permalink] New post 24 Jul 2008, 07:35
Thanks gang :-) I see my mistake.
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Re: remainder [#permalink] New post 24 Jul 2008, 11:13
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?
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Re: remainder [#permalink] New post 24 Jul 2008, 23:06
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
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Re: remainder [#permalink] New post 24 Jul 2008, 23:19
@ 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.
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Re: remainder [#permalink] New post 24 Jul 2008, 23:34
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.
Re: remainder   [#permalink] 24 Jul 2008, 23:34
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