Here’s a look back at Monday’s numeric entry challenge question.
The square root of integer x is equal to the sum of the cubes of y and z. If the squares of y and z are each less than 10, what is the greatest possible value of x?
Answer and Explanation
This is a translation problem that is mighty difficult, but is not necessarily more difficult than something that may appear these days on the GRE. That’s right: the GRE math is getting harder.
Speaking of difficulty level, the Brain Twisters are actually getting easier. Not by too much, mind you, but enough so that they are about at the same level of the most difficult problem on the GRE.
Now back to the problem. Let’s translate the information:
√x = y^3 + z^3
y^2 < 10 z^2 < 10 At this point, we are looking for the maximum value of ‘x’. So if we assume that ‘y’ and ‘z’ equal √10, we will get a number that is one integer higher than the greatest possible value of ‘x’. If y equals √10, then y^3 = 10√10. The same applies to z^3. So going back to the first equation (√x = y^3 + z^3), we get the following: √x = 10√10 + 10√10 √x = 20√10 (√x)^2 = (20√10)^2 x = 400x10 = 4,000 Remember that when we assume that y and z are √10, we are using the lowest possible of y and z that is greater than their actual value. This is the upper bound. In other words, ‘x’ has to be less than 4,000. And the greatest integer less than 4,000 is 3,999. The answer.
Hope you enjoyed the challenge. 🙂 Check back in two weeks for another mind-bending Brain Twister!