# Listen With Others

## Pandigital Squares by Oyler: Solution

Posted by Listen With Others on 16 Jun 2010

From the clue lengths and numbering we can deduce the grid to be as follows.

10d. Q4 is 5 or 6 so Q34 is 25 or 36. So 5a starts with a 5 or 6.

2d. Multiple of 25 or 36 with the multiple being 4, 5, 6 or 9. This yields

 25 36 4 100 144 5 125 180 6 150 216 9 225 324

The only fit is with 150 so Q10 is 6 and Q34 is 25. Entering these in the grid gives up P4 as 2. So we have P as ???2?????? and Q as **25*****6.

8a. Q12 > P12. So Q1 is not 1.

3a. P10 is 4, 5, 6 or 9. But from 8a it is not 4 and from the fact that P4 is 2 it is not 5. So P10 is 6 or 9. If it is 6 then Q12 is 36 but Q10 is 6 so P10 is 9 and 3a is 81. Q1 is 8 and Q2 is 1. Using the 2 digit termini of square numbers tells us that P9 is even but not 2 which already appears in P. Also Q9 is odd and will be 3, 7 or 9.

8d. 6 x 81 = 486 so P6 is 8 and P8 is 6.

12a. P2 or P5 must be 5.

11d. 1 must be in P1 to P4 inclusive.

3d. This is a multiple of 12 so must be 84 and Q8 is 7.

5a. Q6 is 4.

7d. Starts with at least a 4 so 7a must be 90. Therefore Q5 is 9, Q9 is 3 and Q7 is 0. Thus Q is 8125940736 which is 901442.

5a is 5940.

6d ends in 4 so P5 is 4. Now P9 is 0.

13a is ?60.

11d. P2 must be 5.

4d is a multiple of 8 and is 104 so P7 is 7.

12a is 35.

13a is 760.

11d is 30.

1d is 24 so P3 is 3 and P1 is 1.

P is thus 1532487609 which is 391472.

The remaining entries are

7d is 933, 9a is 3248, 1a is 213 and 6d is 914 which check out.

The final grid is as follows: