Listen With Others » Puzzles

Pandigital Squares by Oyler

Listen With Others — Thu, 17 Jun 2010 08:00:13 +0000

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

Listen With Others — Wed, 16 Jun 2010 16:00:10 +0000

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

Listen With Others — Fri, 20 Jun 2008 20:14:00 +0000

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