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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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Updated on: 14 Aug 2009, 06:00
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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

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Originally posted by snipertrader on 14 Aug 2009, 04:20.
Last edited by snipertrader on 14 Aug 2009, 06:00, edited 3 times in total.
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14 Aug 2009, 04:45
1
If n is a positive integer, what is the remainder when 38n+3 + 2 is divided by 5?

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

(38n + 5) MOD 5
=> 38n MOD 5 + 5 MOD 5
=> 35n MOD 5 + 3n MOD 5 + 0
=> 0 + 3n MOD 5
=> 3n MOD 5

Now depending on the value of n, we have:
n = 1 => 3 MOD 5 = 3
n = 2 => 6 MOD 5 = 1
n = 3 => 9 MOD 5 = 4
n = 4 => 12 MOD 5 = 2
n = 5 => 15 MOD 5 = 0

Thus any of the options could be a remainder.
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14 Aug 2009, 04:57
Agree with samrus . Answer could be any of the options depending on the value of n.
if n = 2, remainder will be 1
if n = 3, remainder will be 4,
if n = 0, remainder will be 0.. and so on.
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14 Aug 2009, 05:07
If n is a positive integer, what is the remainder when 38n+3 + 2 is divided by 5?

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

I THINK THERE IS SOMETHING WRONG WITH THE QUESTION PLZ REVISE
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14 Aug 2009, 06:50
Fixed , my mistake
yezz wrote:
If n is a positive integer, what is the remainder when 38n+3 + 2 is divided by 5?

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

I THINK THERE IS SOMETHING WRONG WITH THE QUESTION PLZ REVISE

still remainder can take multiple values it could be any of the answer choices
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Updated on: 14 Aug 2009, 10:23
2
1
Guys, I think this is the question -

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

in which case you will always get a remainder of 4. Plug in and see....hence E is the answer.

Originally posted by sdrandom1 on 14 Aug 2009, 07:57.
Last edited by sdrandom1 on 14 Aug 2009, 10:23, edited 2 times in total.
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Updated on: 19 Aug 2009, 11:12
3
If n is a positive integer, what is the remainder when $$3^(8n+3)$$ + 2 is divided by 5?
Note : (+2 is not raised to 3)

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

SOL:
$$3^{(8n+3)}$$ MOD 5
=> ($$3^{(8n)}$$ * $$3^{3}$$) MOD 5
=> 27 MOD 5 * ($$3^{(8n)}$$) MOD 5
=> 2 * ($$9^{(4n)}$$) MOD 5

Now, 9 MOD 5 = 4 = -1
=> 2 * ($$-1^{(4n)}$$) MOD 5

Now, $$-1^{(Even No)}$$ will always be 1.
=> 2 * 1 MOD 5
=> 2 MOD 5
=> 2

Thus when $$3^{(8n+3)}$$ + 2 is divided by 5, the remainder is 2 + 2 = 4

ANS: E
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Originally posted by samrus98 on 14 Aug 2009, 08:40.
Last edited by samrus98 on 19 Aug 2009, 11:12, edited 1 time in total.
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14 Aug 2009, 09:12
samrus98 wrote:
If n is a positive integer, what is the remainder when $$3^(8n+3)$$ + 2 is divided by 5?
Note : (+2 is not raised to 3)

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

SOL:
$$3^(8n+3)$$ MOD 5
=> ($$3^(8n)$$ * $$3^3$$) MOD 5
=> 27 MOD 5 * ($$3^(8n)$$) MOD 5
=> 2 * ($$9^(4n)$$) MOD 5

Now, 9 MOD 5 = 4 = -1
=> 2 * ($$-1^(4n)$$) MOD 5

Now, $$-1^(Even No)$$ will always be 1.
=> 2 * 1 MOD 5
=> 2 MOD 5
=> 2

Thus when $$3^(8n+3)$$ + 2 is divided by 5, the remainder is 2 + 2 = 4

ANS: E

how come try pluging in 1,2,3,4,5,... for n , we will always get different Remainder values? can you plz shed some light on your way?..i d appreciate
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14 Aug 2009, 09:41
yezz wrote:
samrus98 wrote:
If n is a positive integer, what is the remainder when $$3^(8n+3)$$ + 2 is divided by 5?
Note : (+2 is not raised to 3)

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

SOL:
$$3^(8n+3)$$ MOD 5
=> ($$3^(8n)$$ * $$3^3$$) MOD 5
=> 27 MOD 5 * ($$3^(8n)$$) MOD 5
=> 2 * ($$9^(4n)$$) MOD 5

Now, 9 MOD 5 = 4 = -1
=> 2 * ($$-1^(4n)$$) MOD 5

Now, $$-1^(Even No)$$ will always be 1.
=> 2 * 1 MOD 5
=> 2 MOD 5
=> 2

Thus when $$3^(8n+3)$$ + 2 is divided by 5, the remainder is 2 + 2 = 4

ANS: E

how come try pluging in 1,2,3,4,5,... for n , we will always get different Remainder values? can you plz shed some light on your way?..i d appreciate

Even if you plug in values:
n=1
3^(8 + 3) MOD 5 + 2 MOD 5
= 27*3^8 MOD 5 + 2
= 2 * 9^4 MOD 5 + 2
= 2 * (-1)^4 MOD 5 + 2 ........... (Since 9 MOD 5 = 4 = -1 ie either add 4 or subtract 1)
= 2 * 1 MOD 5 + 2
= 2 + 2
= 4

n=2
3^(16 + 3) MOD 5 + 2 MOD 5
= 27*3^16 MOD 5 + 2
= 2 * 9^8 MOD 5 + 2
= 2 * (-1)^8 MOD 5 + 2
= 2 * 1 MOD 5 + 2
= 2 + 2
= 4
.
..
...
....

n=5
3^(40 + 3) MOD 5 + 2 MOD 5
= 27*3^40 MOD 5 + 2
= 2 * 9^20 MOD 5 + 2
= 2 * (-1)^20 MOD 5 + 2
= 2 * 1 MOD 5 + 2
= 2 + 2
= 4

Hope this helps !!
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14 Aug 2009, 09:51
1
Thanks samrus for taking the time to re-explain to me, that said and as much as my math skills are very basic ( i dont know wut is Mod as i dont have a very sofisticated quantitative back ground as óbviously, your , i just try to plug in numbers in the equation given and devide by 5 and it gives me several values ( there is a celecity). how come your way gives one answer and by plugging in numbers in the equation and deviding does not.
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14 Aug 2009, 10:22
7
yezz wrote:
Thanks samrus for taking the time to re-explain to me, that said and as much as my math skills are very basic ( i dont know wut is Mod as i dont have a very sofisticated quantitative back ground as óbviously, your , i just try to plug in numbers in the equation given and devide by 5 and it gives me several values ( there is a celecity). how come your way gives one answer and by plugging in numbers in the equation and deviding does not.

Hey yezz, no prob.....its just that I aint tht good at explaining.....

MOD function gives the remainder when a number is divided by another number.
Eg. 11 MOD 7 = 4
Any remainder obtained can also be expressed as a negative number.
Eg. 11 MOD 7 = -3
What this means this 11 is 3 short of 7's next multiple which is 14 in this case.

There are some properties associated with MOD:
Property 1: (a*b*c.....) MOD x = (a MOD x) * (b MOD x) * (c MOD x)....
Eg. 11^5 MOD 7 = (11 MOD 7) * (11 MOD 7) * (11 MOD 7) * (11 MOD 7) * (11 MOD 7)
= (4 MOD 7) * (4 MOD 7) * (4 MOD 7) * (4 MOD 7) * (4 MOD 7)

Property 2: (a MOD x) * (b MOD x) * (c MOD x).... = (a*b*c.....) MOD x
Eg. (4 MOD 7) * (4 MOD 7) * (4 MOD 7) * (4 MOD 7) * (4 MOD 7) = (4^5) MOD 7

NOTE: We started out with (11^5 MOD 7) and have reduced it to (4^5 MOD 7). We could have done this directly, since 11 MOD 7 = 4.
Now (4^5 MOD 7) = (2^10 MOD 7)
Doing some rearranging we get,
= (2* 2^9) MOD 7
= (2 MOD 7) * (8^3 MOD 7)
= 2 * (1^3 MOD 7) .......... (Since 8 MOD 7 = 1)
= 2 * 1
= 2

Basically our goal is to reduce the dividend to such a point where it yields a remainder of either 1 or -1 when divided by the divisor. Initially this might seem daunting and slow, but it becomes really fast by practice!

Property 3: (a + b + c.....) MOD x = (a MOD x) + (b MOD x) + (c MOD x)....

Property 4: (a + b*c - d^p) MOD x = (a MOD x) + (b MOD x) * (c MOD x) - (d^p MOD x)

Hope I have been of help!
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14 Aug 2009, 10:46
samrus98 wrote:
yezz wrote:
Thanks samrus for taking the time to re-explain to me, that said and as much as my math skills are very basic ( i dont know wut is Mod as i dont have a very sofisticated quantitative back ground as óbviously, your , i just try to plug in numbers in the equation given and devide by 5 and it gives me several values ( there is a celecity). how come your way gives one answer and by plugging in numbers in the equation and deviding does not.

Hey yezz, no prob.....its just that I aint tht good at explaining.....

MOD function gives the remainder when a number is divided by another number.
Eg. 11 MOD 7 = 4
Any remainder obtained can also be expressed as a negative number.
Eg. 11 MOD 7 = -3
What this means this 11 is 3 short of 7's next multiple which is 14 in this case.

There are some properties associated with MOD:
Property 1: (a*b*c.....) MOD x = (a MOD x) * (b MOD x) * (c MOD x)....
Eg. 11^5 MOD 7 = (11 MOD 7) * (11 MOD 7) * (11 MOD 7) * (11 MOD 7) * (11 MOD 7)
= (4 MOD 7) * (4 MOD 7) * (4 MOD 7) * (4 MOD 7) * (4 MOD 7)

Property 2: (a MOD x) * (b MOD x) * (c MOD x).... = (a*b*c.....) MOD x
Eg. (4 MOD 7) * (4 MOD 7) * (4 MOD 7) * (4 MOD 7) * (4 MOD 7) = (4^5) MOD 7

NOTE: We started out with (11^5 MOD 7) and have reduced it to (4^5 MOD 7). We could have done this directly, since 11 MOD 7 = 4.
Now (4^5 MOD 7) = (2^10 MOD 7)
Doing some rearranging we get,
= (2* 2^9) MOD 7
= (2 MOD 7) * (8^3 MOD 7)
= 2 * (1^3 MOD 7) .......... (Since 8 MOD 7 = 1)
= 2 * 1
= 2

Basically our goal is to reduce the dividend to such a point where it yields a remainder of either 1 or -1 when divided by the divisor. Initially this might seem daunting and slow, but it becomes really fast by practice!

Property 3: (a + b + c.....) MOD x = (a MOD x) + (b MOD x) + (c MOD x)....

Property 4: (a + b*c - d^p) MOD x = (a MOD x) + (b MOD x) * (c MOD x) - (d^p MOD x)

Hope I have been of help!

Thanks man, u tought me a new way...and to be honest... ur great at explaining i appreciate
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Updated on: 14 Aug 2009, 11:00
yezz wrote:
samrus98 wrote:
yezz wrote:
Thanks samrus for taking the time to re-explain to me, that said and as much as my math skills are very basic ( i dont know wut is Mod as i dont have a very sofisticated quantitative back ground as óbviously, your , i just try to plug in numbers in the equation given and devide by 5 and it gives me several values ( there is a celecity). how come your way gives one answer and by plugging in numbers in the equation and deviding does not.

Hey yezz, no prob.....its just that I aint tht good at explaining.....

MOD function gives the remainder when a number is divided by another number.
Eg. 11 MOD 7 = 4
Any remainder obtained can also be expressed as a negative number.
Eg. 11 MOD 7 = -3
What this means this 11 is 3 short of 7's next multiple which is 14 in this case.

There are some properties associated with MOD:
Property 1: (a*b*c.....) MOD x = (a MOD x) * (b MOD x) * (c MOD x)....
Eg. 11^5 MOD 7 = (11 MOD 7) * (11 MOD 7) * (11 MOD 7) * (11 MOD 7) * (11 MOD 7)
= (4 MOD 7) * (4 MOD 7) * (4 MOD 7) * (4 MOD 7) * (4 MOD 7)

Property 2: (a MOD x) * (b MOD x) * (c MOD x).... = (a*b*c.....) MOD x
Eg. (4 MOD 7) * (4 MOD 7) * (4 MOD 7) * (4 MOD 7) * (4 MOD 7) = (4^5) MOD 7

NOTE: We started out with (11^5 MOD 7) and have reduced it to (4^5 MOD 7). We could have done this directly, since 11 MOD 7 = 4.
Now (4^5 MOD 7) = (2^10 MOD 7)
Doing some rearranging we get,
= (2* 2^9) MOD 7
= (2 MOD 7) * (8^3 MOD 7)
= 2 * (1^3 MOD 7) .......... (Since 8 MOD 7 = 1)
= 2 * 1
= 2

Basically our goal is to reduce the dividend to such a point where it yields a remainder of either 1 or -1 when divided by the divisor. Initially this might seem daunting and slow, but it becomes really fast by practice!

Property 3: (a + b + c.....) MOD x = (a MOD x) + (b MOD x) + (c MOD x)....

Property 4: (a + b*c - d^p) MOD x = (a MOD x) + (b MOD x) * (c MOD x) - (d^p MOD x)

Hope I have been of help!

Thanks man, u tought me a new way...and to be honest... ur great at explaining i appreciate

Great work !!!! Kudos ! This is a new method for me as well........but it still doesnt explain the discrepancy between plugging in values and samrus's method rite??

Originally posted by sergbov123 on 14 Aug 2009, 10:51.
Last edited by sergbov123 on 14 Aug 2009, 11:00, edited 1 time in total.
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14 Aug 2009, 10:53
yezz wrote:
Thanks samrus for taking the time to re-explain to me, that said and as much as my math skills are very basic ( i dont know wut is Mod as i dont have a very sofisticated quantitative back ground as óbviously, your , i just try to plug in numbers in the equation given and devide by 5 and it gives me several values ( there is a celecity). how come your way gives one answer and by plugging in numbers in the equation and deviding does not.

Infact it would be great if you could show how exactly you go about plugging in values.....coz samrus's method, although brilliant, looks really complex.....
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14 Aug 2009, 11:04
1
If n is a positive integer, what is the remainder when $$3^(8n+3)$$ + 2 is divided by 5?
Note : (+2 is not raised to 3)

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

so the question is 3^(8n+3) + 2 not 3(8n+3)+2

if so then,

3^0dd , lets test remainders

3^1 , r= 3
3^3,r = 2

3^5,r = 3

3^7 , r = 2,

3^9,r= 3... we have a cilicity when 3 is raised to an odd power.

3^11,r=2, 3^13,r= 2 ( if n=1), if n = 2 then we are looking at 3^19,r= 2, 3^27,r= 2 as well

thus 3^(8n+3)/5 is always 2

thus when you add this 2to the 2 from (3^(8n+3) + 2) you get 4 as a remainder
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14 Aug 2009, 19:23
Great post samrus98 +1 i think we need a turbo kudos button for posts like these.

Hope that GMAT has none or only few questions like these!!

yezz - nicely done..thnx
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21 Aug 2009, 07:52
I got the remainder of 4 in the following way:
3^(8n+3)+2=3^8n×3^3+2=27×3^8n+2=(25+2)×3^8n+2=25×3^8n+2×3^8n+2
25×3^8n is divisible by 5 and a remainder is 0.
So the remainder will be equal to the remainder of the following:
2×3^8n+2=2×(3^8n+1)=2×(9^4n+1)
observing that 9^4 has a units digit of 1, ((9^4)^n+1) has a units digit of 2 and the expression 2×(9^4n+1) has a unit digit of 4.
a two digit number with a a units digit of 4 can be presented as 10k+4. since 10k is divisible by 5, the remainder will be 4.
a three digit number can be written in the form 100m+10k +4. Since 100m+10k is divisible by 5, the remainder is 4...
etc...
any number with units digit of 4 will have a remainder of 4, when it is divided by 5.
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21 Aug 2009, 11:09
for this one, we just put n=1 in the equation

(3^11+2)=9^5*3+2
the last digit of this number is 9
so the remainder must be 4.

Don't try to sort out the whole equation, it can't be done within 2 minutes.
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22 Aug 2009, 07:46
samrus98 wrote:
yezz wrote:
Thanks samrus for taking the time to re-explain to me, that said and as much as my math skills are very basic ( i dont know wut is Mod as i dont have a very sofisticated quantitative back ground as óbviously, your , i just try to plug in numbers in the equation given and devide by 5 and it gives me several values ( there is a celecity). how come your way gives one answer and by plugging in numbers in the equation and deviding does not.

Hey yezz, no prob.....its just that I aint tht good at explaining.....

MOD function gives the remainder when a number is divided by another number.
Eg. 11 MOD 7 = 4
Any remainder obtained can also be expressed as a negative number.
Eg. 11 MOD 7 = -3
What this means this 11 is 3 short of 7's next multiple which is 14 in this case.

There are some properties associated with MOD:
Property 1: (a*b*c.....) MOD x = (a MOD x) * (b MOD x) * (c MOD x)....
Eg. 11^5 MOD 7 = (11 MOD 7) * (11 MOD 7) * (11 MOD 7) * (11 MOD 7) * (11 MOD 7)
= (4 MOD 7) * (4 MOD 7) * (4 MOD 7) * (4 MOD 7) * (4 MOD 7)

Property 2: (a MOD x) * (b MOD x) * (c MOD x).... = (a*b*c.....) MOD x
Eg. (4 MOD 7) * (4 MOD 7) * (4 MOD 7) * (4 MOD 7) * (4 MOD 7) = (4^5) MOD 7

NOTE: We started out with (11^5 MOD 7) and have reduced it to (4^5 MOD 7). We could have done this directly, since 11 MOD 7 = 4.
Now (4^5 MOD 7) = (2^10 MOD 7)
Doing some rearranging we get,
= (2* 2^9) MOD 7
= (2 MOD 7) * (8^3 MOD 7)
= 2 * (1^3 MOD 7) .......... (Since 8 MOD 7 = 1)
= 2 * 1
= 2

Basically our goal is to reduce the dividend to such a point where it yields a remainder of either 1 or -1 when divided by the divisor. Initially this might seem daunting and slow, but it becomes really fast by practice!

Property 3: (a + b + c.....) MOD x = (a MOD x) + (b MOD x) + (c MOD x)....

Property 4: (a + b*c - d^p) MOD x = (a MOD x) + (b MOD x) * (c MOD x) - (d^p MOD x)

Hope I have been of help!

samrus98, simply superb.
even though orginal prob does not require this method to solve it. The above explained method was an eye-opener.
But, I would like to correct some info in your post, as this post will be reference for learning this property to many ppl.

Quote:
NOTE: We started out with (11^5 MOD 7) and have reduced it to (4^5 MOD 7). We could have done this directly, since 11 MOD 7 = 4.
Now (4^5 MOD 7) = (2^10 MOD 7)
Doing some rearranging we get,4
= (2* 2^9) MOD 7
= (2 MOD 7) * (8^3 MOD 7)
= 2 * (1^3 MOD 7) .......... (Since 8 MOD 7 = 1)[/strike]
= 2 * 1
= 2

I have question in the usage of exponents here,
2^9 can be written as 2^(3*3)
(2^3)*(2^3) = 64

or it can be written as,
2^(3^2) = 8^2 = 64
but, you have written as,
2^9 = 2^(3*3) = 8^3 = 512
Whether, writing, 2^9 = 8^3 is correct?
But, it does not have any problem since the base is 1 here in the subsequent steps.
Please, Let us know the steps/properties to consider while simplifying the exponents.
Please correct me if I am wrong.

regards,
hhk
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22 Aug 2009, 10:33
1
hariharakarthi wrote:
I have question in the usage of exponents here,
2^9 can be written as 2^(3*3)
(2^3)*(2^3) = 64

or it can be written as,
2^(3^2) = 8^2 = 64
but, you have written as,
2^9 = 2^(3*3) = 8^3 = 512
Whether, writing, 2^9 = 8^3 is correct?
But, it does not have any problem since the base is 1 here in the subsequent steps.
Please, Let us know the steps/properties to consider while simplifying the exponents.
Please correct me if I am wrong.

regards,
hhk

It would help to take note of the following:
1) 2^3 * 2^4 = 2^(3 + 4) = 2^7
2) (2^3)^4 = 2^12 = 8^4
3) 2^3^4 = 2^81

2^(3^2) = 8^2 = 64 is not correct!!
2^(3^2) = 2^9 = (2*2*2)*(2*2*2)*(2*2*2) = 8^3

2^9 = 2^(3*3) is fine, but 2^(3*3) is not equal to (2^3)*(2^3)!!

I hope this will help.....
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Re: Remainder Question &nbs [#permalink] 22 Aug 2009, 10:33

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