# Archive for the ‘Puzzles’ Category

## Pandigital Squares by Oyler

Posted by Listen With Others on 17 June 2010

Ten double sided cards each have a different single digit printed on each side. When the cards are arranged in a row a pandigital square, P, is formed. When the cards are turned over and kept in the same order the result is a different pandigital square Q. In the clues the subscripts refer to the cards in positions 1 to 10 respectively. For example if P was 6154873209 then P25 would be the four digit string 1548. In order for solvers to identify P and Q, the grid, which has 180° rotational symmetry, should be completed. In the grid no entry starts with zero and all are different. P and Q should be written underneath the grid.

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 Across Down 1 P13 + P89 1 P3 x P6 3 P10 x P10 = Q12 2 Q10 x Q34 5 Q47 3 Q8 x Q23 7 Q3 x Q4 x Q5 4 P6 (P7 + P8) 8 P4 (Q12 – P12 ) / Q9 6 P46 + Q46 + P34 + Q67 9 P36 7 P24 + Q68 – Q10 12 P2 x P7 8 Q10 x Q12 13 P79 10 Q4 x Q4 = Q34 11 P1 x P2 x P3 x P4

## 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. 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:

## Zero-less Pandigital Products: Prime 234 by Oyler

Posted by Listen With Others on 20 June 2008

In a first for Listen With Others, the following is a puzzle by Oyler that was sent to all solvers who submitted feedback with their solution to the puzzle.

Many thanks to Oyler for allowing this to be published here, and I hope that people have as much fun solving it as I did. If anybody has any feedback on this puzzle, then I will happily pass it on to Oyler, or it can be posted as comments at the bottom of this posting.

Chris

Zero-less Pandigital Products : Prime 234

By Oyler

In the clues P, Q and R are a two digit, three digit and four digit number respectively that between them contain all of the digits from 1 to 9 with P prime such that PQR = X_Y_Z where X_Y_Z is a nine digit number split into three groups of three digits that again contains all of the digits from 1 to 9.  For example 29 x 871 x 5364 = 135,489,276.  Across entries are in capitals and X¢ denotes the reverse of X.  All the entries are different, the normal rules of algebra apply and there are no zeros in the grid.

Clues

 P Q R X Y Z A dp 9A2 2q f cd K + g G 4Cm 4D ad 8D K + j G¢ p¢q – F – N – P N 6h 9C M¢ F2 FHc J 6B 4b q¢ FH 7 ( k – n ) E¢ ce + c A¢H