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

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Intern
Joined: 13 Jul 2009
Posts: 19

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22 Aug 2009, 16:39
samrus98, thanks a lot.

I found where I made mistake,
2^9 = 2^(3*3)
from here on we should simplify as, (2^3)^3 which is equal to 8^3.
But, i tried to write 2^9 = 2^(3^2), then simplified as, 8^2 which is wrong.

Exponents, we should simplify from right to left; top to bottom.

Lessons learnt. Again Thanks a lot for helping me in identifying the error. Kudos to U..

regards,
hhk.
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Joined: 10 Aug 2009
Posts: 129

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22 Aug 2009, 17:01
hariharakarthi 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!

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

2^9=8^3 is correct.
2^3*2^3 does not equal 2^9. 2^3*2^3=2^(3+3)=2^6 or 2^3*2^3=(2X2)^3=4^3
http://www.platinumgmat.com/gmat_study_ ... tions#laws

I am sorry I did not notice that there was a second page and samrus98 has already replied. Anyway, the link might be helpful.
Senior Manager
Joined: 29 Jul 2009
Posts: 284

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Updated on: 23 Aug 2009, 09:23
4
I have another method to solve this problem, I think nobody has posted it.

We want to know the reminder that a number yields when it is divided by 5. If we knew the last digit of that number the answer would be easy.

now what is the last digit of 3^(8n+3)?

the pattern for the powers of the number 3 is --> 3,9,7,1
this pattern has a length of 4 digits
so dividing 8n + 3 by 4 the reminder is 3. This operation tells us that the last unit digits of 3^(8n+3) is 7

now 7 + 2 makes 9, the last digit number of 3^(8n+3) + 2; therefore the reminder that we're looking for is 4

Originally posted by mikeCoolBoy on 23 Aug 2009, 04:22.
Last edited by mikeCoolBoy on 23 Aug 2009, 09:23, edited 1 time in total.
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Joined: 25 Jul 2009
Posts: 114
Schools: NYU, NUS, ISB, DUKE, ROSS, DARDEN

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23 Aug 2009, 08:29
mikeCoolBoy wrote:
I have another method to solve this problem, I think nobody has posted it.

We want to know the reminder that a number yields when it is divided by 5. If we knew the last digit of that number the answer would be easy.

now what is the last digit of 3^(8n+3)?

the pattern for the powers of the number 3 is --> 3,9,7,1
this pattern has a length of 4 digits
so dividing 8n + 3 by 4 the reminder is 3. This operation tell us that the last unit digits of 3^(8n+3) is 7

now 7 + 2 makes 9 the last digit number of 3^(8n+3) + 2; therefore the reminder that we're looking for is 4

Brilliant !! +1 for you!
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Manager
Joined: 28 Jul 2009
Posts: 116
Location: India
Schools: NUS, NTU, SMU, AGSM, Melbourne School of Business

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24 Aug 2009, 03:58
mikeCoolBoy wrote:
I have another method to solve this problem, I think nobody has posted it.

We want to know the reminder that a number yields when it is divided by 5. If we knew the last digit of that number the answer would be easy.

now what is the last digit of 3^(8n+3)?

the pattern for the powers of the number 3 is --> 3,9,7,1
this pattern has a length of 4 digits
so dividing 8n + 3 by 4 the reminder is 3. This operation tells us that the last unit digits of 3^(8n+3) is 7

now 7 + 2 makes 9, the last digit number of 3^(8n+3) + 2; therefore the reminder that we're looking for is 4

That is how I figured out. Good one mike!
+1 for you.
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Manager
Joined: 17 Aug 2009
Posts: 196

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09 Dec 2009, 10:35
I got this wrong initially when i attempted it as I didn't 2/5 as well

Now see,

3 ^ (8n+3) will always have 1 as the last digit (cyclicity of 3----3,9,7,1)

Therefore since it is given that it is a positive integer, we need to start from 1

3^11/5 +2/5 will give remainder (1+3) = 4 always

try this for 3^19 (n=2) and 3^27 (n=3) and you will get the same result
Senior Manager
Joined: 25 Jun 2009
Posts: 256

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09 Dec 2009, 21:07
Would they consider this question a 700+ level question?
Manager
Joined: 05 Jun 2009
Posts: 73

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09 Dec 2009, 21:12
I hope so, I actually know how to do remainder exponent problems...
Senior Manager
Joined: 25 Jun 2009
Posts: 256

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09 Dec 2009, 22:12
It took me a little bit, but I finally gained a much better understanding of it from the MGMAT Number Properties book.
Senior Manager
Joined: 22 Dec 2009
Posts: 317

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31 Jan 2010, 05:29
samrus98 wrote:

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 a lot for this Sameer...... Do u recommend any books to make such fundamentals strong??? I wasn't even aware of this concept!
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Joined: 18 Feb 2010
Posts: 160
Schools: ISB

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05 Mar 2010, 01:11
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

3^ (8n+3) = 3^8n * 3^3

we know 3 has a cyclicity of 4 with last digit as 1 (3^4 = 81)

hence from above we know 3^8n has last digit as 1 coz 8n is a mutilple of 4

Now with second part we know 3^3 is 27 with remainder 2 when divided by 5

so remainders combined 1*2 + 2 =4

4/5 remainder is 4 answer E

Trust me it may look big but the solution takes a maximum of 1 minute.

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