Finding what a value is least divisible by

[IBstudent]

I am having alot of trouble with this question, I would really appreciate it if someone could help me with it...

A and B are non zero integers. If 2A⁴-3AB=0, then B is divisible AT LEAST by...
Answer choices: 2, 3, 9, 18.

My way of solving it was substituting values into the equation, I then noticed that all of the values of B were divisble by 2. But that is not the answer.

Please help :S

 

[Deleted member 4993]

I am having alot of trouble with this question, I would really appreciate it if someone could help me with it...

A and B are non zero integers. If 2A⁴-3AB=0, then B is divisible AT LEAST by...
Answer choices: 2, 3, 9, 18.

My way of solving it was substituting values into the equation, I then noticed that all of the values of B were divisble by 2. But that is not the answer.

Please help :S

2A⁴-3AB=0 → A*(2A³-3B)= 0

If A = 0, we cannot say anything about B

If A \(\displaystyle \ne\) 0 → 2A³-3B = 0 → B has to be an even number

Now continue....

 

[IBstudent]

2A⁴-3AB=0 → A*(2A³-3B)= 0

If A = 0, we cannot say anything about B

If A \(\displaystyle \ne\) 0 → 2A³-3B = 0 → B has to be an even number

Now continue....

Okay, so 2A³ = 3B.
Which means that B was to be even. However, most of the answers are even. Hmmmm, but since we want the LEAST value that B is divisble by, the answer should be 2. But its not!!!!

The answer should be 18.

Please show me what I did wrong

 

[Yogi]

My way of solving it was substituting values into the equation,

What were these values you substituted and their results?

 

[srmichael]

Okay, so 2A³ = 3B.
Which means that B was to be even. However, most of the answers are even. Hmmmm, but since we want the LEAST value that B is divisble by, the answer should be 2. But its not!!!!

The answer should be 18.

Please show me what I did wrong

\(\displaystyle A^3=\frac{3B}{2}\)

thus, B must be a value that produces a perfect cube since A must be an integer. The only even integer of the choices that B can be is 18.

\(\displaystyle A^3=\frac{(3)(18)}{2}\)

\(\displaystyle A^3=\frac{54}{2}\)

\(\displaystyle A^3=27\) ===> Perfect Cube!

 

[Deleted member 4993]

I don't like the way the term "at least" is used in this question. Terrible choice of word...

I would have formulated the question following way:

Find the LARGEST possible divisor of B, among the set of numbers {2, 3, 9, 18} such that 2A⁴ - 3AB = 0, where A and be are positive non-zero integers.

 

[IBstudent]

\(\displaystyle A^3=\frac{3B}{2}\)

thus, B must be a value that produces a perfect cube since A must be an integer. The only even integer of the choices that B can be is 18.

\(\displaystyle A^3=\frac{(3)(18)}{2}\)

\(\displaystyle A^3=\frac{54}{2}\)

\(\displaystyle A^3=27\) ===> Perfect Cube!

Thanks! I finally got it!!!

And I do agree that the question is vague/confusing.