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Is A a number between 40 and 90 to give a 6-digit number?

A will produce a four digit number, the other two cells being taken up by an addition (see preamble).
Good luck!

The eleven letters stand for the numbers 1 - 11. You then have to insert a number between 1 and 11 into each result to get the grid entry.

I hate these numericals, but I always give them a go. A is immediate, but there are two possible numbers to insert into 1A. Putting the lower one with the zero in the only place leads me quickly to an impossibility, so it must be the other double digit. I can't find the next step in the logic though. It would appear from posts here that it's fairly easy, so I must be missing something obvious.

Cockie - your analysis so far is correct. Nothing else as immediate as 1a, but 14a and 30d will tell you 4 letters have to be 4 numbers (but not which is which). 19d then tells you another 2 (again but not which is which). Hope that gets you going.

Just starting, but I have a question: if the insertion is 10 or 11 does one digit or both digits of the insertion have to be in (a) checked square(s), or is there complete freedom? It would help in pinning down possibilities for 1 across.

throck, the second digit must be in a checked square, otherwise there would not be a unique solution. The first digit could be unchecked. In other words, the message is not used to eliminate possible alternative solutions.

Help needed (ashamed...but this is from the start!!)

- I deduce from 14a and 30d that H, N, S and E produce at most 2 digit cubes, ie all are from {1,2,3,4}.

- 13a tells me that I produces, at most a 4 digit fourth power, ie is {1,2}

So five letters from 4 digits is not possible.....what am I missing?

your first deduction is correct

I is in fact 5 (IIII would then be 625 )

The first part is correct, jezzac. I don’t follow the second part. As smellyharry mentioned earlier, look at 19dn to narrow down I and G. That should put you right on 13ac