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If n[i] and m are positive integers, what is the remainder

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chrism
 Post subject: If n[i] and m are positive integers, what is the remainder
 Post Posted: Fri Aug 24, 2007 9:34 pm 
If n[i] and m are positive integers, what is the remainder when "3^(4n+2) + m" is divided by 10 ?

(1) n = 2
(2) m = 1

The answer is 'B', but I don't get it!!! if m is one, you still don't know what 3^(4n+2) is... right? all we know is it's a power of 3... so it's units digit could be any number between 0-9.... thus, we still don't know what the remainder would be if divided by 10.... please help!!!!


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chrism
 Post subject:
 Post Posted: Fri Aug 24, 2007 9:38 pm 
Image


here's the cut&paste of the problem
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GMAT 2007
 Post subject:
 Post Posted: Fri Aug 24, 2007 10:05 pm 
Rephrase the expression: -
3^(4n+2) +m = (9)*3^(4n) + m

statement (1) n =2 so the expression = (9)*3^8 + m but we do not know what m is - so cannot predict the value of the expression. INSUFFICIENT

statemtn (2) m = 1 which makes the expression:

(9)*3^(4n) + 1 Since we know n is +ve integer, now it gets tricky: -

for n = 1,2,3,4 the exponential component of the expression will be

3^4, 3^8, 3^12 or
9^2, 9^4, 9^6 or

81, 81^2, 81^3 ans so on... the unit digit of all these values will be 1, now this value will be multiplied by 9 and '1' will be added to the result. It will make the unit digit of the result - 0. It means the result will be prefectly divided by 10. So the remainder will be 0

SUFFICIENT, So the answer is (B)

Hope it helps

GMAT 2007
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StaceyKoprince
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 Post Posted: Sat Sep 01, 2007 6:39 pm 
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Very nice, GMAT 2007!

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grge.jcb
 Post subject: Re: If n[i] and m are positive integers, what is the remainder
 Post Posted: Tue Jun 02, 2009 1:32 pm 
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i have an easier method
3^1 =3 ,3^2=9,3^3=27,3^4 =81,3^5 = 243
notice a pattern in the units digit? they repeat every fourth power. 3^1 & 3^5 have the same units digit,3^2 & 3 ^6 have the same units digit & so on
using (1)
9*3^4n + m becomes 9*3^8 + m
considering only units digit , 9*1 + m
INSUFFICENT
using (2)
9*3^4n + 1 , as shown above, for all values of n, units digit 3^4n remains the same. ( UD of 3^4=1,UD of 3^8 =1)
Now , considering only units digit
9*1 + 1 = 10 ,Hence B SUFFICIENT
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RonPurewal
 Post subject: Re: If n[i] and m are positive integers, what is the remainder
 Post Posted: Mon Jun 15, 2009 10:09 pm 
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Posts: 7146
grge.jcb wrote:
i have an easier method
3^1 =3 ,3^2=9,3^3=27,3^4 =81,3^5 = 243
notice a pattern in the units digit? they repeat every fourth power. 3^1 & 3^5 have the same units digit,3^2 & 3 ^6 have the same units digit & so on


this is good.

in fact, this is an excellent takeaway.

TAKEAWAY:

in REMAINDER PROBLEMS:
if you don't INSTANTLY see the algebraic solution, then IMMEDIATELY start LOOKING FOR A PATTERN.

there's also a fact that you should know concerning this problem statement:

fact:
REMAINDERS UPON DIVISION BY 10 are simply UNITS DIGITS.

for instance, when 352 is divided by 10, the remainder is 2.

since remainders are fundamentally based on stuff repeating over and over and over again, it shouldn't be a surprise that patterns emerge early and often among remainders.

this solution isn't necessarily "easier" - that judgment depends upon how comfortable you are with the algebra and theory - but it can be quite efficient.

Quote:
using (1)
9*3^4n + m becomes 9*3^8 + m
considering only units digit , 9*1 + m
INSUFFICENT


this is a good analysis, but you don't even need to try that hard.

instead, you can just realize that, since m can be anything at all, you can have any units digit you want.

Quote:
using (2)
9*3^4n + 1 , as shown above, for all values of n, units digit 3^4n remains the same. ( UD of 3^4=1,UD of 3^8 =1)
Now , considering only units digit
9*1 + 1 = 10 ,Hence B SUFFICIENT


yeah.

technically, you should also throw away the "1" in your sum of 10, reducing to a final units digit of 0.

but well played.
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