Question

Let x, y and z be three positive integers satisfying \(y = 3x\), \(z = 4x\), and \(x + y + z = 3k\), where k is an integer. Which of the following is the smallest value of k for which x, y and z are even numbers?

• A. 8
• B. 10
• C. 12
• D. 16

Hint:

Write x + y + z as 8x and find when 8x/3 is a whole number, then add the requirement that x itself be even, since y = 3x needs x even while z = 4x is automatically even.

Answer 1

The Correct Option is D

Step 1: Write the sum in terms of x.
We are given \(y = 3x\) and \(z = 4x\). Adding all three:
\[ x + y + z = x + 3x + 4x = 8x \]
We are told this sum equals \(3k\), so \(8x = 3k\), which gives \(k = \dfrac{8x}{3}\).

Step 2: Find which values of x make k an integer.
Since 8 and 3 share no common factor, \(\dfrac{8x}{3}\) is a whole number only when x itself is a multiple of 3. So x can be 3, 6, 9, 12, and so on.

Step 3: Add the condition that x, y, z must all be even.
\(y = 3x\) is even exactly when x is even, since multiplying by the odd number 3 never changes whether a number is odd or even. \(z = 4x\) is always even no matter what x is, because 4 is even. So the only real restriction from y and z is that x itself must be even.
Combining this with Step 2, x must be a multiple of 3 AND even, which means x must be a multiple of 6.

Step 4: Take the smallest valid x.
The smallest positive multiple of 6 is \(x = 6\). Check: \(y = 3(6) = 18\), \(z = 4(6) = 24\). All of 6, 18 and 24 are even, so this works.

Step 5: Work out k.
\[ k = \frac{8x}{3} = \frac{8(6)}{3} = \frac{48}{3} = 16 \]

Step 6: Check the other options are too small.
For k to be smaller, x would have to be a smaller multiple of 6, which is not possible since 6 is already the smallest positive one. Trying x = 3 (a multiple of 3 but not of 6) gives y = 9 and z = 12; y is odd, so this fails the all even condition even though it gives a smaller k = 8. So options (1) 8, (2) 10 and (3) 12 do not correspond to a valid set of all even x, y, z.

Final Answer:
The smallest value of k for which x, y and z are all even is 16. \[ \boxed{k = 16} \]