Any suggestions? Need to put brackets in to this true. Pls help! @mathsjem @solvemymaths @hegartymaths pic.twitter.com/LOdAAYOYqE

— Japleen Kaur (@japleen_kaur1) May 14, 2015

Any ideas…?

Apparently this is a “mastery task” for Year 7 students (11-12). A bit of a cruel one if so.

Placing brackets into an expression is just about establishing the order in which the various operations will be performed. Wherever you put the brackets, one of those four operations, **×** , **−** , **+** or **×** again, will have to be performed LAST. Which one could it be?

Well, it can’t be either of the multiplications. If your final operation is **2 × SOMETHING** or **SOMETHING × 6**, then your answer will definitely be **even**. We want the answer to be **13**, an **odd** number, so that’s no good.

The last operation can’t be the **−** either, because that would mean that we have already evaluated the first multiplication and got **2 × 3 = 6**. That only leaves even numbers in our calculation, and there is no way of putting even numbers together with only **−** , **+** and **×** to make an odd number.

The last operation must be the **+** then, right? If so, that means that we have already evaluated the second multiplication to get **2 × 6 = 12**, so we have **SOMETHING + 12 = 13**. Clearly the SOMETHING is **1**, so the question becomes how to make **2 × 3 − 2 = 1**. Unfortunately, whether we do the multiplication first: **(2 × 3) − 2 = 4**; or the subtraction first: **2 × (3 − 2) = 2**; it can’t be done.

Or have I missed something…?

Oh, and while we’re on the subject of arithmetic failures, don’t spiders have eight legs?