Data Sufficiency

BTGmoderatorLU wrote:Source: Veritas Prep

\(W, X, Y\), and \(Z\) represent distinct digits such that \(WX*YZ=1995\). What is the value of \(W\)?

1) \(X\) is a prime number.
2) \(Z\) is not a prime number.

The OA is D

Given:

1. \(W, X, Y\), and \(Z\) represent distinct digits
2. \(WX*YZ=1995\).

We have to get the value of W.

Since the units digit of 1995 is 5, the units digit of the product of X*Z must also be 5; thus, there are three possibilities: 1. X = 1 and Z = 5; 2. X = 5 and Z = 1; 3. X = 5 and Z = 5. The third possibility does not exist since X and Y are distinct integers.

Let's take each statement one by one.

1) \(X\) is a prime number.

=> X = 5 and Z = 1

=> W5 * Y1 = 1995. Since 5*1 = 5 (a single digit; no carry-over), upon multipilcation, the units digit of (w + 5y) must be the tens digit of 1995, i,e, 9.

From W5 * Y1 = 1995, we have (10W + 5)*(10Y + 1) = 1995 => 100WY + 50Y + 10W + 5 = 1995 => 10WY + 5Y + W = 199.

Since the units digit of 10WY would be 0, the units digit of (5Y + W) would be 9.

Case 1: Say Y = 2; thus, 5y = 10 and W must be 9. Plugging in the values of Y and W, we get

10WY + 5Y + W = 199 => 10*9*2 + 5*2 + 9 = 199 => 180 + 10 + 9 = 199 = 199. Satisfies.

Case 2: Say Y = 3; thus, 5y = 15 and W must be 4. Plugging in the values of Y and W, we get

10WY + 5Y + W = 199 => 10*4*3 + 5*3 + 4 = 199 => 120 + 15 + 4 = 139 ≠ 199. Does not satisfy.

Note that Y cannot be 1 or 5 since X = 5 and Z = 1. We know that the digits are distinct. If Y = 4/6/8, W = 9, Same value that we calculated in Case 1.

Case 3: Say Y = 7; thus, 5y = 35 and W must be 4. Plugging in the values of Y and W, we get

10WY + 5Y + W = 199 => 10*4*7 + 5*7 + 4 = 199 => 280 + 35 + 4 ≠ 199. Does not satisfy.

There is no need to check for Y = 9 since at higher value, 10WY + 5Y + W >> 199.

Thus, W = 9. Sufficient.

2) \(Z\) is not a prime number.

=> Z = 1 and X = 5. It is the same result that we got in Statement 1. Sufficient.

The correct answer: D

Hope this helps!

-Jay
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