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

We can also solve this by thinking about odd and even behaviour directly, instead of jumping straight to the algebra.

We know $y = 3x$ and $z = 4x$, and $x + y + z = 8x = 3k$.

1. Why z is always even: $z = 4x$ is a multiple of 4, and any multiple of 4 is automatically even, no matter what x is.
2. Why y's evenness depends only on x: $y = 3x$. Multiplying by the odd number 3 never changes whether a number is odd or even, so y is even only when x is even, and y is odd only when x is odd.
3. Why x itself must be even: since we need x, y and z all even, and y is even only when x is even, x has to be even. There is no way around this.
4. Why x must also be a multiple of 3: from $8x = 3k$, the right side is a multiple of 3, so the left side, $8x$, must be too. Since 8 has no factor of 3, x itself must supply it, so x must be a multiple of 3.

Putting the two requirements together, x must be even AND a multiple of 3, which forces x to be a multiple of 6. The smallest positive multiple of 6 is $x = 6$.

With $x = 6$: $y = 18$, $z = 24$. All three, 6, 18 and 24, are even, exactly as required.

Now find k from the sum: $x + y + z = 6 + 18 + 24 = 48$, and $48 = 3k$ gives $k = 16$.

Let's summarize:

• z is always even, no matter what x is.
• y is even only when x is even, so x must be even.
• k is a whole number only when x is a multiple of 3.
• Both conditions together force x to be a multiple of 6, and the smallest such x is 6, giving k = 16.

So the smallest value of k for which x, y and z are all even is 16.