Challenge Question from 6/01/09

[alexambroz]

Here's the 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

And the official answer from MGMAT:

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.
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."

The correct answer is B: Statement (2) is sufficient, but Statement (1) is not.

My question is: in the bold part for Statement (2) --> why do we know that a+1-c cannot be less than zero? The question says that they are all positive integers, but why couldn't 'a' be equal to some random pos integer and c be some other random pos integer (say, a = 3 and c = 4,000). Thanks.

-AA

[esledge]

Note for clarity: in the problem above, "2c" notation means 2^c or "2 to the c power."

My question is: in the bold part for Statement (2) --> why do we know that a+1-c cannot be less than zero? The question says that they are all positive integers, but why couldn't 'a' be equal to some random pos integer and c be some other random pos integer (say, a = 3 and c = 4,000). Thanks.

-AA

The answer lies in the constraint for p. If p = 2m/n = [2^(a+1-c)]*[3^b], neither a+1-c nor b can be negative, lest p be a fraction. This provides a redudant constraint for b, which we already know is positive, but provides a new constraint on the relationship between the positive integers a and c.