Beautiful, very tricky math problem

[laiusergiu]

a, x, and k are positive integers. k is odd. Is a divisible by 24?

(1) x^(2a) + 1/[x^(2a)]>2
(2) x^[(k^3)-k] = x^a

[laiusergiu]

The answer is C: Both statements 1 and 2 together are sufficient to answer the question but neither statement is sufficient alone. See if you can figure it out.

[galiya.dautova]

The answer is C: Both statements 1 and 2 together are sufficient to answer the question but neither statement is sufficient alone. See if you can figure it out.

This must be an error - either on your part (transcription) or on the part of Manhattan GMAT. If a is odd, clearly it cannot be divisible by 24.

if its possible could you send me the source please? i mean where MGMAT explains the decision

[jnelson0612]

laiusergiu, what is the original source of this question?

[hoxlay]

a, x, and k are positive integers. k is odd. Is a divisible by 24?

(1) x^(2a) + 1/[x^(2a)]>2
(2) x^[(k^3)-k] = x^a

What's the source of the question?

Anyway, here goes a solution:

Question is basically asking whether both 3 AND 8 are factors of 'a' or in other words, is 'a' a multiple of 24?

1) For LHS to be greater than 2, x must be greater than 1, since when x=1, LHS = 2, and 2 can not be greater than itself. Thus all positive integer values of x, such that x >1 satisfy this statement. And this makes all positive integer values of 'a' suitable to satisfy the statement.

Thus statement 1 merely tell us that is an integer such that a>0. Statement 1 is NOT SUFFICIENT (to determine whether 'a' is a multiple of 24.)


2)since the two sides of the equation have the same base, we can equate the exponents ===>

(k^3)-k] = a
k[(k^2)-1] = a
k[(k^2)-(1^2)] = a
k(k-1)(k+1) = a

Now, 3 is ALWAYS a divisor/factor of 'a' since 'a' is product of three consecutive (doesn't matter whether positive or negative) integers. So, since k is odd, (k-1)(k)(k+1) is always divisible by 8 since 'a' is zero when k is 1, or 'a' is a multiple of 8 when k is an odd number greater than 1. You can confirm this by plugging in odd integers 3 and 5.

Basically for any positive odd integer k, 'a' is a product of three integers, two of which are different even integers, i.e. k-1 (always even) and k+1 (also always even and different from k-1), hence 'a' is always divisible by 8.

Thus 'a' is divisible by both 3 AND 8, and hence 'a' is a multiple of 24.

when k=3, a= (2)(3)(4)= 24
when k=5, a =(4)(5)

Statement 2 is SUFFICIENT.

Answer is B

[laiusergiu]

Hi everyone,

Sorry to respond with a delay.

hoxlay... you did a great job with 2 of the subtleties of this problem, but got tripped up by the third. I'll explain.

Statement 1: You were perfectly right that the only thing that statement one tells us is that x is not equal to one. That may seem irrelevant, but it isn't. You'll see why in a moment.

Statement 2: hoxlay, this is where you got tripped up. You cannot equate the exponents just because the two sides have the same base. Not until you've made sure that the base is not equal to {-1, 0, 1}. Consider the following three scenarios:
1^8=1^20...but obviously 8 does not equal 20
(-1)^8=(-1)^20...but obviously 8 does not equl 20
0^8=0^20...but obviously 8 does not equl 20

We know from the conditions of the problem that x is a positive integer, and thus we know that x cannot equal {-1,0}. However, x may equal 1, as a result we cannot equate the exponents. We thus conclude that Statement 2 is insufficient.

When we put the two statements together, we know that x cannot equal 1 either, and therefore we can proceed to equate the exponents of the two sides in Statement 2. From there, hoxlay, your divisibility explanation is perfect and shows us why we have sufficient information to conclude that a is a multiple of 24.

I hope you enjoyed it. I composed the problem myself.
Thanks,
Sergiu

[hoxlay]

Great question. Thanks for pointing out the final subtlety. Cheers.

[jnelson0612]

We still need the original source of the question, if only to confirm this isn't a real GMAT question. I suspect that it is not, but need to ask.

[laiusergiu]

Jamie,
I stated above that I composed the problem myself.
Thanks

[jnelson0612]

Good to know, thanks!