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Source: — Data Sufficiency |
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by codesnooker » Thu May 01, 2008 3:04 am
Solution:

Equation = 3^(4n+2) + m

It can be reduced as

= 3^4n + 3^2 + m
= 3^4n + 9 + m

According to 2nd statement, m = 1

= 3^4n + 9 + 1
= 3^4n + 10

Now if you can check,
3^0 = 1
3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^8 = 81 * 81 = 6561

So, it means no matter what is the value of n (0, 1, 2, ....) the last digit would be always 1.
Now if you add 10 this number, the unit digit will always be 1. So, when the number would be divided by 10, then the reminder would always be 1.

Hence, statement (2) is alone sufficient to deduce the answer.
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by gmatjoe » Thu May 01, 2008 4:09 am
Thanks codesnooker, I agree with your calculation.
However, how did you arrive at 3^((4n+2)+m)?
I cannot make this out of the question stem, I read that the n could be n as in stat1 or m as in stat2...
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by codesnooker » Thu May 01, 2008 5:06 am
Unbelievable!!!

The question is misprinted. Same question I solved yesterday that I found at this forum.

Here is the link of correct question statement.

https://www.beatthegmat.com/gmat-prep-qu ... 10282.html

Check it and you can understand it yourself.
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by gmatjoe » Thu May 01, 2008 6:21 am
thanks, it's indeed unbelievable. i already started to doubt about my reading skills. browsing around, this is already the second misprint i encountered.
anyone knows of gmac updating gmatprep once and a while?
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by gmatjoe » Thu May 01, 2008 6:32 am
I opened a thread in the gmat math general folder to collect these kinds of errors...

https://www.beatthegmat.com/misprints-in ... html#41767
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by onlyGmat2008 » Sat May 24, 2008 12:12 am
Hi codesnooker,

How can Equation = 3^(4n+2) + m be reduced as 3^4n + 3^2 + m.

shoouldn't it be 3^(a+b) is 3^a . 3^b . Please correct if i am misinterpreting something.

Also i didn't understand the final solution where you say that the last digit is always 1. Could you please help me understand this?

Thanks.
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by gmataug08 » Fri Jun 13, 2008 10:13 am
the equation : 3^(4n+2) + m

with considering the second data information m = 1

3^(4n+2) + 1 => 3^4nx3^2 + 1 => 3^4nx9 + 1

now considering 3^4n
3^4 = 81 , hence whatever the value of n it would be so many times multiples of 81.
for ex :
when n =1 ==> 3^(4x1) ==> 3^4 ==> 3x3x3x3 =>81
when n =2 ==> 3^(4x2) ==> 3^8 ==> 3x3x3x3x3x3x3x3 => 81x81
when n =3 ==> 3^(4x3) ==> 3^12 ==> 81x81x81

so, irrespective of n's value the unit digit is always going to be 1

now back in equation ,
3^4nx9 would always have a unit digit of 9 ( as the unit digit is going to be 1 always for 3^4n) .... adding 1 to that value would make the unit digit to be always 0.

so when dividing this number by 10 , the reminder will always be 0.
so the second equation alone enough .

hence answer B.
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