Bunuel wrote:
(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?
(1) x = y - 6
(2) x + y+ z = 13
100x+10y+z = three-digit integer xyz
10x+y = two-digit integer xy
Since dividing xyz by 5 yields integer 10w+z, xyz must be a MULTIPLE OF 5.
xyz will be a multiple of 5 only if nonzero digit z is 5.
Plugging z=5 into (100x +10y + z) ÷ 5 = 10w + z, we get:
\(\frac{100x + 10y + 5}{5} = 10w + 5\)
\(100x + 10y + 5 = 50w + 25\)
\(100x + 10y = 50w + 20\)
\(10x + y = 5w + 2\)
The resulting equation implies the following:
Two-digit integer xy is equal to 2 more than a multiple of 5, with the result that
y=2 or y=7.
Since w≤9, two-digit integer xy is less than or equal to 47, with the result that
x=1, x=2, x=3 or x=4.
Statement 1, rephrased: y = x+6
Here, only x=1 and y=7 satisfy the blue constraints above.
Since the values of x and y are known, the value of w can be determined.
SUFFICIENT.
Statement 2, rephrased with z=5: x+y = 8
Here, only x=1 and y=7 satisfy the blue constraints above.
Since the values of x and y are known, the value of w can be determined.
SUFFICIENT.
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