If x, y, z, w are all positive integers less than 10 that satisfy the

(100x +10y + z) ÷ 5 = 10w + z
If x, y, z, w are all positive integers less than 10 that satisfy the above equation, what is the value of w?

\(20x +2y + \frac{z}{5} = 10w + z\)
\(20x +2y - \frac{4}{5}z = 10w\)

So, z = 0 OR z = 5 (multiple of 5)
For z = 0, y = 1,2,3,.. depending on values of x and w
For z = 5, y = 2 (w = 2, x =1 e.g.)
or
z = 5, y = 7 (w = 3, x = 1 e.g.)

(1) x = y - 6
\(20(y - 6) +2y - \frac{4}{5}z = 10w\)
\(22y -120 - \frac{4}{5}z = 10w\)
\(22y - \frac{4}{5}z - 10w = 120\)
z ≠ 0 as it would lead to an equation that is not possible with the conditions given(22y is a multiple of 10 only when y = 5 which is not possible).
For z = 5,
\(22y - 10w = 124\)
22 * 7 - 10 * 3 = 124 (only one solution)
w = 3

SUFFICIENT.

(2) x + y+ z = 13
\(20(13 - y - z) +2y - \frac{4}{5}z = 10w\)
\(260 - 20y - 20z +2y - \frac{4}{5}z = 10w\)
\(-18y - \frac{104}{5}z - 10w = -260\)
\(18y + \frac{104}{5}z + 10w = 260\)
Again, z ≠ 0 as it would lead to an equation that is not possible with the conditions given('18y +10w' is not a multiple of 10).
For z = 5,

\(18y + 10w = 260 - 104\)
\(18y + 10w = 156\)
18 * 7 + 10 * 3 = 156 (taking a clue from St. 1 OR from unit digit of 156)
w = 3

SUFFICIENT.

Answer D.

(100x +10y + z) ÷ 5 = 10w + z

Or, 2x + y/5 - 2*z/25 = w, it is important to note, x, y, z and w are positive integers less than 10.
So, z= 5 and y= 7. Now, we need value of x only to get value of w.

Stat1: x = y - 6
Yes! we can get value of x, so, value of W is possible. Sufficient.

Stat2: x + y+ z = 13
Yes! we can get value of x, so, value of W is possible. Sufficient.

So, I think D.

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