**Kenken Hints: Finding hidden “impossibles” that allow you to eliminate a possibility**

Ok, here’s a new **combination of boxes**I found that **defines **other boxes. It happens to be a 7×7, but the point is the **principle** that it illuminates. I’ve often found this tactic helpful once in mid-game, where the obvious fills have been done and now something more hidden must be found. It must be used carefully, because it’s easy to fool yourself that things are determined when they aren’t.

So here’s the example:

At first glance both the **1-**’s could be anything. (I leave **1-**‘s blank or put in a very faint “A” for “all” rather than putting in all the numbers until I have reduced the possibilities). **6x **can **only **be **1 & 6 **or **2 & 3**. If **6x -> 1 & 6, then 6+ ( **which is a single value here from the pairs, **1 + 5 **and

**2 + 4) **must be either **2** or **3**. If **6x -> 2 + 4, **then **6+ **must be either **2, 3, **or **4. **But my “unknown” **1-’**s, of course, can’t have any number the same. So **1 + 6 leads to an impossible condition, **namely no two pairs can fit between 1 & 6 if any other number between 1 & 6 is required in the

**6+**box. So

**6x must be 2 & 3**to allow the

**1-’s**to be

**4 & 5**and

**6 & 7**.

At first glance both the **1-**’s could be anything. (I leave **1-**‘s blank or put in a very faint “A” for “all” rather than putting in all the numbers until I have reduced the possibilities). **6x **can **only **be **1 & 6 **or **2 & 3**. If **6x -> 1 & 6, then 6+ ( **which is a single value here from the pairs, **1 + 5 **and

**2 + 4) **must be either **2** or **3**. If **6x -> 2 + 4, **then **6+ **must be either **2, 3, **or **4. **But my “unknown” **1-’**s, of course, can’t have any number the same. So **1 + 6 leads to an impossible condition, **namely no two pairs can fit between 1 & 6 if any other number between 1 & 6 is required in the

**6+**box. So

**6x,**

*to allow the*must be 2 & 3!**1-’s**to be**4 & 5**and**6 & 7**.