How balanced do you think the puzzle types of this test were?	Perfectly balanced
What was your opinion of the distribution of easy/hard puzzles?	Just right
What did you think about the puzzle quality of the test?	Very nice
How did you feel about the length / time limit for this test?	Way too many puzzles (too little time)
Of the puzzles you solved/attempted, how well did the point values reflect the difficulty?	Most puzzles were worth the right amount
What was your opinion of the booklet formatting and printing?	Just right

Can the password please be removed? I didn't have a chance to compete in this contest, but still would like to solve the puzzles. Thanks!

Will the scores in this test be included in the ratings

It will be removed. For now: RS7x7=49

Have fun!

Yes, but it may take some time before it is done.

Thanks Richard! They were fun. I do like the 7x7 format!

Attempted the test . Just loved it ..... but now want to solve the puzzles at leisure - PW to the PUZZLE BOOKLET PLEASE !

Removed the password.

thanks...
Will try to go thru the Solving tips now

Am working slowly thru the 7*7 Sudoku tips.
Vide Quad sums . A. could not understand R7C1=4; R1C7 = 4 ( 4 in C1 R 6 and 7 , similarly 4 in C 7 R 1 and 2 ) . B. How was 2 ruled out in R67C6 ( you mentioned 2 in C7 in R67; 3 in R6 in C67 ) - 3 is understood in R6C67 .
Vide Consequtive . R2 contains 2 dots which must have pairs {12} and {45} . How were pairs 56 , 67 ruled out ? Then next how was [ R2C13 = {67}; R13C2 = {15} ] derived ?

I only saw vjain's questions today, so here is a belated response:



There is a 4 in R3C2. Because of that, the 4 in R1 must be somewhere in R1C3-7. Those cells are all in the same irregular area, so the 4 is ‘locked’ there. Hence, the 4 is not in R23C7. There is also a 4 in R5C6; because of being in that shape, no 4 in R4567C7. The only possibility for the 4 in C7 is now R1C7. This works exactly the same for R7C1.




The 2 in C7 must be in the irregular area bottom right, more specific, in R4567C7. The sum-dot in that irregular area must contain the series 1-2-3-6 since the 4 is already placed elsewhere in that area. Combining both previous conclusions, the 2 in that irregular area goes in R67C7.



There is a given 3 in row 2 without a dot next to it. This means that the 2 and the 4 in row 2 have to go in two of the four remaining cells in that row. There are two dots there, both connecting two different cells. The 2 and 4 cannot be placed next to the same dot, so this means that in those 4 cells you find the pairs {12} and {45}, without knowing which pair goes where at this stage.
The 6 and 7 in row 2 have to go in R2C13.
Since the 2 and 4 also cannot be placed above or below the given 3 (no dots there too), the 1 and 5 have to go there.


Hope this helps, and sorry for the delay. Must be my holiday that I didn't notice your questions before!