Bunuel wrote:
If the square root of the product of three distinct positive integers is equal to the largest of the three numbers, what is the product of the two smaller numbers?
(1) The largest number of the three distinct numbers is 12.
(2) The average (arithmetic mean) of the three numbers is 20/3.
Bunuel, is this a problem you created, or is it from an existing source? It's a cleverly designed question. One can use trial and error to show that Statement 2 is also sufficient, but it's not required, as shown below:
Let our numbers be a, b, c, where 0 < a < b < c. Then if sqrt(abc) = c, it must be that ab = c, so to find ab (and thus answer the question) we only need to find the largest of our three numbers, and Statement 1 is sufficient.
For Statement 2, if the mean is 20/3, then the sum of our three numbers is 20, so we know: a + b + ab = 20, and a + b + ab is certainly even. If a and b were both odd, or if one were odd and the other even, then a + b + ab would be odd, which is not the case. So a and b must both be even. Thus a and b are two different positive even numbers which give a product less than 20. If a were greater than 2, then a would be at least 4 and b would be at least 6 (since b > a), which gives too large a product. So a must be 2, and using the equation above, substituting a=2, we have 2 + b + 2b = 20, or 3b = 18, and b = 6.
So Statement 2 not only lets us find the value of ab, it actually lets us find both a and b individually, and is sufficient, and the answer is D.
I took this question from some blog. And the OA given was A. But I disagreed. I think the answer should be D.
(1) Clearly sufficient.
\(a+1=1\) and \(b+1=21\), doesn't work, as \(a\) becomes 0 and we know that integers are more than 0.
\(a+1=3\) and \(b+1=7\). \(a=2\) and \(b=6\) --> \(ab=12\)
Hence sufficient.