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Ones digit Challenge

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Ones digit Challenge

by Jim@StratusPrep » Sun Jul 08, 2012 4:58 pm
For the positive integers W, X, Y, and Z, Z = 2x + 1, and W=Y^Z. What is the remainder of W when it is divided by 10?

1) Y = 9
2) X = 3
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Source: — Data Sufficiency |
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by eagleeye » Sun Jul 08, 2012 5:16 pm
Jim@StratusPrep wrote:For the positive integers W, X, Y, and Z, Z = 2x + 1, and W=Y^Z. What is the remainder of W when it is divided by 10?

1) Y = 9
2) X = 3
The key to solving this lies in recognizing that Z=2X+1 means that Z is odd.
Also, when a positive integer is divided by 10, the remainder is the units digit itself.
After that the problem is a breeze. We are told that W=Y^2 and we need to find whether we can determine the units digit of W.

1) Y=9
W=9^Z, but recall that Z is odd.
Now 9 is a cool number because 9^1=9 , 9^2 =9*9 = 81, 9^3= 81*9 =729, 9^4=729*9 = something ending in 1. We see that for even
powers of 9, last digit is 1, for odd powers it is always 9. (The pattern practically shouts itself).
Since the units digit of W for odd Z is always 9, this is Sufficient.

2) X=3, this means that Z=7.
For Y=1, W=1^7 = 1, units digit is 1
For Y=2, W=2^7 = 128, units digit is 8
(I wrote these counter examples for reference only, in reality if we have an expression W^7, it is obvious that the units digit is going to depend on W, and hence we can't determine what it is!)
Indeterminable units digit, Hence Insufficient.

A is the final answer.
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