remainder?

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remainder?

by Ahmed MS » Wed Jun 15, 2011 11:52 pm
If n and m are positive integers, what is remainder when 34n + 2 + m is divided by 10?

a. n = 2
b. m = 1


The answer is B, but I failed to understand! Please guys help me!

Cheers!
Source: — Data Sufficiency |

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by Geva@EconomistGMAT » Thu Jun 16, 2011 12:32 am
Ahmed MS wrote:If n and m are positive integers, what is remainder when 34n + 2 + m is divided by 10?

a. n = 2
b. m = 1


The answer is B, but I failed to understand! Please guys help me!

Cheers!
I know the question - it actually asks for 3^(4n + 2) + m

When a number is divided by 10, the remainder is the units digit of the number. 53 divided by 10 gives a remainder of 3. likewise, 7724 divided by 10 will give a remainder of 4.

So the question is really asking "what is the units digit of 3^(4n + 2) +m?

Stat. (2) is sufficient because powers of 3 have a recurring pattern of units digits:

3^1 = 3
3^2 = 9
3^3 = 27 ends with a 7
3^4 = 81 ends with a 1
3^5 = ...take that 1, multiply by 3, and you get something that ends with a 3.
3^6 = ...take the 3 from the previous 3^5, multiply times another 3, and get something that ends with a 9.
3^7 = 9*3 = 27, so ends with a 7.
3^8 = 7*3 = 21, so ends with a 1.

Etc. etc.
The pattern of units digits repeats every four places: 3, 9, 7, 1.

That's why the fact that 3is raised to the power of 4n is significant: for every value of n, 3^4n will be a power with a multiple of 4: 3^4, 3^8, 3^12, etc. All of these powers will end with the same units digit of 1, for whatever value of n. If the power is 4n+2, then the units digit is fixed at "2 more than the nearest multiple of 4", 3^6, 3^10, 3^14, etc. which will always have a units digit of 9.

So in order to determine the value of the units digit if 3^(4n+2) +m, all we really need to know is the value of m.
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by Ahmed MS » Thu Jun 16, 2011 1:43 am
Dear Geva@MasterGMAT,


Thanks for your reply. Actually, I got the questions from web. There were errors in the questions.

Cheers!