A few months ago, I wrote a couple of articles targeted toward those students looking for a super-high score (one for quant, one for verbal). I challenged students to answer those questions in much less time than we typically average on test questions.

Well, I’m back with another one in the series. This problem is a bit different though: it’s from our Challenge Problem archive, a question bank consisting of what we call 800+ level problems. (Some might qualify as 750+ but most are harder than anything you’ll ever see on the real test.)

Do you need to be able to answer a question like this in order to score 750+? Absolutely not. (In fact, after my colleague Ron Purewal submitted this question, I tested it out on several of my fellow instructors, all of whom have scored 760+ on the test. Not everyone answered correctly.) Mostly, I’m offering this to stretch your brains, drive you a little crazy, and make one important point (see my second takeaway at the end).

If, however, quant is your strength and you’re hoping to score 51 in that section”you can certainly score 51 without getting this one right, but if you *do* get this one right in 2 minutes, then you know you’re ready for the quant section.

One more tidbit before we dive in. I chose this question because it is SO very hard. As of right now (as I’m typing this), 254 people have tried this problem and 44 have answered it correctly.

Do a little math here. What percentage of people answered the question correctly?

17%. Random guess position is 20%. Wow.

All right, enough with the build up. Are you ready to try? The below problem is copyright ManhattanGMAT, originally published in April 2013. Kudos to Ron for thinking up this devilish problem!

* If

xis positive, what is the value of |x– 3| – 2|x– 4| + 2|x– 6| – |x– 7|?(1)

xis an odd integer(2)

x> 6

I’m not listing the five data sufficiency answer choices here. If you don’t already have them memorized, then put this article away and come back to it when you’re further along in your studies.

Okay, so what did you get? I’m going to share two solutions with you. One is the official solution published with the problem. The other is the way that I did the problem when I first saw it. My solution method is faster and mostly only involves thinking about the problem”but I chose not to write that up as the official solution because I think most people will have trouble following it, even those who are going for a 51 on quant. I have more room here, though, so I’ll show you that method too.

Here’s the official solution:

(1) SUFFICIENT: It’s impractical to take an algebraic approach to this statement; doing so would entail a large number of cases. For instance, |

x– 3| is equal to 3 –xifx< 3, but is equal tox- 3 ifx> 3; similarly, the other three absolute-value expressions switch atx= 4, 6, and 7, respectively.Instead, it’s more efficient to consider the first three positive odd integers (1, 3, and 5) individually and then to consider only one algebraic case, the case in which

x>7 (because whenx> 7, all values are positive so we can ignore the absolute value symbols).If

x= 1, then the value is (2) “ 2(3) + 2(5) “ (6) = 0.

Ifx= 3, then the value is (0) “ 2(1) + 2(3) “ (4) = 0.

Ifx= 5, then the value is (2) “ 2(1) + 2(1) “ (2) = 0.If

x> 7, then drop the absolute value symbols and simplify:

(x– 3) “ 2(x– 4) + 2(x– 6) “ (x– 7) =

x– 3 – 2x+ 8 + 2x– 12 –x+ 7 =

(x– 2x+ 2x–x) + (-3 + 8 – 12 + 7) =0Therefore, the value of the expression is also 0 for all values of

xgreater than or equal to 7, including the odd integers 7, 9, 11, and so on. The value of the expression is thus 0 for all positive odd integers. The statement is sufficient.(2) NOT SUFFICIENT: As determined during the discussion of statement 1, the expression is equal to 0 when

x> 7(whether odd integer, even integer, or non-integer). We still need to test the non-integer values ofxbetween 6 and 7.If

x= 6.5, then the value is (3.5) – 2(2.5) + 2(0.5) – 0.5 = “1, which is not equal to 0. The expression can thus have multiple values, so the statement is insufficient.The correct answer is A.

Wow. Some serious work there. Take your time and go over it carefully”it might help to write out the steps yourself. You’ll find it easier to understand my thinking solution below if you understand how things worked above. (Though you still won’t find it easy!)

Alright, are you ready? The first thing I noticed was some interesting symmetry in the expression |*x* – 3| – 2|*x* – 4| + 2|*x* – 6| – |*x* – 7|. The problem uses four terms but skips the middle term *x* – 5. The other terms are all symmetrical around this middle term: *x* – 4 is one more and *x* – 6 is one less, for example.

Further, the pairs we’re discussing have constant relationships. The term *x* – 3 is always 4 larger than *x* – 7. This four-units-apart relationship will always be true, no matter the value of *x*.

Likewise, the term *x* – 4 is 2 larger than *x* – 6; these two terms are always 2 units apart. Also note that, in the expression given in the problem, these two terms are *both* multiplied by 2. This multiplication will actually preserve the symmetry and the two terms will be 4 apart rather than 2 apart.

Finally, notice the relationships between the addition and subtraction in the given expression. We add the largest term (*x* – 3) and subtract the smallest (*x* – 7). Conversely, of the other pair, we add the smaller term (*x* – 6) and subtract the larger (*x* – 4). In other words, even though the problem at first makes everything positive using those absolute value symbols, it then turns two of the terms negative again (by subtracting them instead of adding them).

Basically, *everything’s* symmetrical! Wait, so does that mean that the value will always be zero no matter what? Not quite (if that were the case, we wouldn’t need any statements at all to answer the question and that’s not how data sufficiency works). Everything looks symmetrical so far, but there’s still some key info missing.

Let’s go back to that missing middle term: *x* – 5. If *x* equals 5, then this term would equal 0. The values of the two terms *x* – 3 and *x* – 7 would be symmetrical about zero (one positive and one negative). Likewise, the two terms *x* – 4 and *x* – 6 would be symmetrical about zero (one positive and one negative). (Again, the absolute value signs make all of these values positive at first, but then the *x* – 4 and *x* – 7 terms turn negative again.)

Not sure about the above case? Write out the real numbers to see how it works.

With a starting point of *x* = 5, then, the value of the expression has to be zero. This won’t work the same way for every possible value for *x*, though”those absolute value symbols do mess things up. We need symmetry.

Either the terms need to balance perfectly (in the case *x* = 5), or the values of the four terms need to be all non-negative (0 or positive) or all non-positive (0 or negative). The four values can’t cross the positive / negative barrier (except in the one perfectly balanced case, *x* = 5) or the symmetry will be messed up.

The problem told us *x* is positive. Any value of *x* < 3 would make all of the terms (*x* – 3, *x* – 4, *x* – 6, and *x* – 7) zero or negative. Therefore, those ones will return a final result of zero.

For 3 < *x* < 7, the only number that provides perfect symmetry is *x* = 5. The others mess up the symmetry and therefore won’t provide a final value of zero.

For all *x* > 7, all of the terms (*x* – 3, *x* – 4, *x* – 6, and *x* – 7) will be zero or positive. Therefore, those ones will also return a final result of zero.

Take a look at the statements. Statement 1 says that *x* is odd. We know that all odd integers 3 and below will result in a final value of 0. The same is true for 5 and for all odd integers 7 and above. Statement 1 is sufficient.

Statement 2 says that *x* > 6. Anything 7 or above will result in a final value of zero”but this statement doesn’t say that *x* is an integer! Any decimal between 6 and 7 will not result in a final value of zero. Statement 2 is not sufficient.

Whew. We’re done. If you find that second explanation way too crazy to follow, don’t worry about it. In fact, even if you think you’d never get the first solution, that’s okay, too. As I said, this problem is harder than anything you’d see on the real test”it’s one of the hardest in our Challenge Problem pool, and every question there is already really hard!

**Key Takeaways**

(1) You don’t really need a 99^{th} percentile score. (Well not unless you want to work for us!) I’m offering the above in the spirit of fun and intellectual curiosity; please don’t feel that you really need to know this for the GMAT!

(2) If you are going for a super-high score, the take-away isn’t that you need to be able to do problems just like this one. You should, though, be able to think or theorize your way through *some* problems”in a similar fashion to what I did in the second solution above, but for easier (though still hard!) problems.

* Challenge Problem ©ManhattanGMAT 2013

Just corrected a typo in the article – at one point, the explanation referred to statement 2 as x 6.

Thanks to Jason Lee for alerting me!