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by skalevar » Sun Jul 18, 2010 1:41 pm
If n and m are positive integers what is the remainder when 3^(4n +2) + m is divided by 10?

1) n = 2
2) m =1

Official answer is B, 2) along is sufficient but I don't understand why. Any takers?
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by sk818020 » Sun Jul 18, 2010 4:02 pm
First you'll notice a pattern;

3^1=3
3^2=9
3^3=27
3^4=81
3^5=243
3^6=729
3^7=2187
3^8=6561
3^10=59049

Some other thing to note;

3^(4n+2)
- 3 will always in this example be raised to a positive power. So you should tell yourself to look for patterns in the even powers.
- The power that 3 will be raised to will always be 2 or a multiple of 4, plus two. So, 2, 6=[4(1)+2], 10=[4(2)+2] and so forth. Note this and then look at the numbers above. 2, 6, and 10 all end in 9. When you notice a pattern like this you need to take advantage of it.

Going off the notion that 3^(4n+2) will have a nine in it's digits place we only need to know what the value of m is to solve the question.

(2) tells us what m is so we can solve the problem. Thus, 2 is sufficient.
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by Patrick_GMATFix » Mon Jul 19, 2010 3:12 pm
skalevar wrote:If n and m are positive integers what is the remainder when 3^(4n +2) + m is divided by 10?

1) n = 2
2) m =1
Hi Skalevar,

The key is to solving this Q quickly is to come up with a strong rephrase. Recognize 2 general rules:

1) remainder when an integer is divided by 10 is the units digit of that integer. So the question can be rephrased as: What is the units digit of 3^(4n+2) + m?

2) units digit of an exponential expression follows a cyclical pattern as the exponent is incremented

In this case rephrasing is key. Since 3^(4n+2) is 9^(2n+1). In other words, it is 9 raised to an odd exponent. Regardless of the exponent, the units digit of 9^(odd) is 1. So to determine the units digit of 3^(4n+2) + m, we only need the units digit of m.

Rephrase: What is the units digit of m?

Statement 2 gives us this info so it's sufficient. The answer is B.

A more detailed explanation and a step-by-step video solution is available at GMATPrep Question 1395. If you struggle with advanced remainder problems, set topic='Number Properties' and difficulty='700+' in the Drill Engine.

Good luck,
-Patrick
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by deeyah » Thu Jul 22, 2010 8:30 pm
Patrick_GMATFix wrote:
skalevar wrote:If n and m are positive integers what is the remainder when 3^(4n +2) + m is divided by 10?

1) n = 2
2) m =1
Hi Skalevar,

The key is to solving this Q quickly is to come up with a strong rephrase. Recognize 2 general rules:

1) remainder when an integer is divided by 10 is the units digit of that integer. So the question can be rephrased as: What is the units digit of 3^(4n+2) + m?

2) units digit of an exponential expression follows a cyclical pattern as the exponent is incremented

In this case rephrasing is key. Since 3^(4n+2) is 9^(2n+1). In other words, it is 9 raised to an odd exponent. Regardless of the exponent, the units digit of 9^(odd) is 1. So to determine the units digit of 3^(4n+2) + m, we only need the units digit of m.

Rephrase: What is the units digit of m?

Statement 2 gives us this info so it's sufficient. The answer is B.

A more detailed explanation and a step-by-step video solution is available at GMATPrep Question 1395. If you struggle with advanced remainder problems, set topic='Number Properties' and difficulty='700+' in the Drill Engine.

Good luck,
Patrick

PLease explain. Regardless of the exponent, the units digit of 9^(odd) is 1. So to determine the units digit of 3^(4n+2) + m, we only need the units digit of m..
Regards
Deeyah
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