PLEASE REPLY - Challenge Problem 6-1-09

[vasd.inc]

PLEASE HELP! In the explanation given for the following problem, MGMAT states that a + 1 - c cannot be less than zero. I can't figure out why that is... why can't c be greater than a+1?

Question:

Positive integers a, b, c, m, n, and p are defined as follows: m = 2a 3b, n = 2c, and p = 2m/n. Is p odd?

(1) a < b
(2) a < c


Answer:

We should first combine the expressions for m, n, and p to get the following:
p = 2m/n = 2(2a 3b) / 2c = 2a + 1 - c 3b

The question can be rephrased as "Does p have no 2's in its prime factorization?" Since p is an integer, we know that the power of 2 in the expression for p above cannot be less than zero (otherwise, p would be a fraction). So we can focus on the exponent of 2 in the expression for p: "Is a + 1 - c = 0?" In other words, "Is a + 1 = c?"

Statement (1): INSUFFICIENT. The given inequality does not contain any information about c.

Statement (2): SUFFICIENT. We are told that a is less than c. We also know that a and c are both integers (given) and that a + 1 - c cannot be less than zero[/color].
In other words, a + 1 cannot be less than c, so a + 1 is greater than or equal to c. The only way for a to be less than c AND for a + 1 to be greater than or equal to c, given that both variables are integers, is for a + 1 to equal c. No other possibility works. Therefore, we have answered our rephrased question "Yes."

[esledge]

Just to be clear, some of those variables are exponents, which I will indicate with the ^ mark. Here's the full question again:

Positive integers a, b, c, m, n, and p are defined as follows: m = (2^a)(3^b), n = 2^c, and p = 2m/n. Is p odd?

(1) a < b
(2) a < c

If p = 2m/n, substituting the given expressions for m and n we get:

p = 2*(2^a)(3^b)/(2^c) = (2^1)(2^a)(3^b)/(2^c) = [2^(1+a-c)]*(3^b)

If 1+a-c were any integer less than zero, the power of two would be a fraction. For example, if 1+a-c = -1 and b = 1, then p = (2^-1)*(3^1) = 3/2. Since we are told that p is an integer, 1+a-c cannot be negative.

[cariappa.88]

Does the question answer itself?

a, b, c, n, p are all positive integers.

m = 2^a * 3^b

Here, a^2 has to be even, and 3^b has to be odd irrespective of what the values of a or b are. As a result, m has to be even.

n = 2^c

Here, n has to be even

p = 2m/n or m = 2(m/n). Here (m/n) could be odd or even, but when multiplied by 2, it becomes even. So, doesn't the question answer itself.

Simply put, any positive integer multiplied by 2 is even. So, 2(m/n) has to be positive.

I'd really appreciate it if someone thinks this explanation is faulty.

Thanks!

[jnelson0612]

Does the question answer itself?

a, b, c, n, p are all positive integers.

m = 2^a * 3^b

Here, a^2 has to be even, and 3^b has to be odd irrespective of what the values of a or b are. As a result, m has to be even.

n = 2^c

Here, n has to be even

p = 2m/n or m = 2(m/n). Here (m/n) could be odd or even, but when multiplied by 2, it becomes even. So, doesn't the question answer itself.

Simply put, any positive integer multiplied by 2 is even. So, 2(m/n) has to be positive.

I'd really appreciate it if someone thinks this explanation is faulty.

Thanks!

Be careful! What if m=6 and n=4? Then you get 3 as the result. Even though m itself is even, if the n contains lots of twos those twos may cancel out with the coefficient 2 up to and also the two contained in the m.