## Pandigital Squares by Oyler: Solution

Posted by Listen With Others on 16 June 2010

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

10d. Q_{4} is 5 or 6 so Q_{34} 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 Q_{10} is 6 and Q_{34} is 25. Entering these in the grid gives up P_{4} as 2. So we have P as ???2?????? and Q as **25*****6.

8a. Q_{12} > P_{12}. So Q_{1} is not 1.

3a. P_{10} is 4, 5, 6 or 9. But from 8a it is not 4 and from the fact that P_{4} is 2 it is not 5. So P_{10} is 6 or 9. If it is 6 then Q_{12} is 36 but Q_{10} is 6 so P_{10} is 9 and 3a is 81. Q_{1} is 8 and Q_{2} is 1. Using the 2 digit termini of square numbers tells us that P_{9} is even but not 2 which already appears in P. Also Q_{9} is odd and will be 3, 7 or 9.

8d. 6 x 81 = 486 so P_{6} is 8 and P_{8} is 6.

12a. P_{2} or P_{5} must be 5.

11d. 1 must be in P_{1} to P_{4} inclusive.

3d. This is a multiple of 12 so must be 84 and Q_{8} is 7.

5a. Q_{6} is 4.

7d. Starts with at least a 4 so 7a must be 90. Therefore Q_{5} is 9, Q_{9} is 3 and Q_{7} is 0. Thus Q is 8125940736 which is 90144^{2}.

5a is 5940.

6d ends in 4 so P_{5} is 4. Now P_{9} is 0.

13a is ?60.

11d. P_{2} must be 5.

4d is a multiple of 8 and is 104 so P_{7} is 7.

12a is 35.

13a is 760.

11d is 30.

1d is 24 so P_{3} is 3 and P_{1} is 1.

P is thus 1532487609 which is 39147^{2}.

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:

## Leave a Reply