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The sum of three numbers is 40 . One number is five more than a second number. It is also twice the third. Find the numbers.

The smallest of the numbers is $\square$.

Step 1 ：Let's denote the three numbers as x, y, and z.

Step 2 ：From the problem, we have the following three equations: \(x + y + z = 40\), \(x = y + 5\), and \(x = 2z\).

Step 3 ：First, let's substitute equation \(x = y + 5\) into equation \(x + y + z = 40\): \((y + 5) + y + z = 40\), which simplifies to \(2y + z + 5 = 40\), and further simplifies to \(2y + z = 35\).

Step 4 ：Next, let's substitute equation \(x = 2z\) into the equation we just got: \(2y + (x/2) = 35\), which simplifies to \(2y + (2z/2) = 35\), and further simplifies to \(2y + z = 35\). This is the same as the equation we got before, so it checks out.

Step 5 ：Now, let's substitute equation \(x = 2z\) into equation \(x = y + 5\): \(2z = y + 5\), which simplifies to \(y = 2z - 5\).

Step 6 ：Substitute \(y = 2z - 5\) into \(2y + z = 35\): \(2(2z - 5) + z = 35\), which simplifies to \(4z - 10 + z = 35\), and further simplifies to \(5z - 10 = 35\), and finally simplifies to \(5z = 45\), so \(z = 9\).

Step 7 ：Substitute \(z = 9\) into \(y = 2z - 5\): \(y = 2*9 - 5\), which simplifies to \(y = 18 - 5\), so \(y = 13\).

Step 8 ：Substitute \(y = 13\) and \(z = 9\) into \(x + y + z = 40\): \(x + 13 + 9 = 40\), which simplifies to \(x = 40 - 13 - 9\), so \(x = 18\).

Step 9 ：So, the three numbers are 18, 13, and 9. The smallest of the numbers is \(\boxed{9}\).