Listen With Others

Are you sitting comfortably? Then we’ll begin

3984 – Odd One Out by Oyler

Posted by Listen With Others on 20 June 2008

Saturday 31st May, 11am A bit of a late start this week what with certain commitments but good to see the return of Oyler, a Listener regular since 1996. He gave us Dedication last August, which was a fairly straightforward tribute to Euler. At first glance, Odd One Out looks far from straightforward but there is a lot of information in the preamble on the third grid (III) and it is immediately clear that the seven Across entries in order will have the lengths: 2, 3, 4, 5, 4, 3, 2 enabling the positioning of the vertical bars as follows:

 
 
[Later I discovered that there was enough information in the preamble to also enter the six horizontal bars at this early stage.]
 
I first look at the two clues:
 
d = 5G : A / b : B + g
A = Cube : Fibonacci : Prime
d = 5G is too big for grids I and II so must belong to grid III where it has 2 or 3 digits.
d = A / b can only belong to grid II and A(II) must be a Cube or Fibonacci number.
 
Now considering:
 
b = AB : Fibonacci : 2E
b(II) can only be Fibonacci or 2E (AB(II) is too big), giving:
{A(II), b(II), d(II)} = {5832, 54, 108} as the sole fit where b is 2E (b even & >19) or a Fibonacci number.
b(II) = 54 = 2E(II)
E(II) = 27
 
Now looking at the two across clues, A again and G, specifically the value A / 3 in G where A cannot be Prime:
 
A = Cube(II) : Fibonacci : Prime
G = Prime : A / 3 : b / A
We know the value of G = A / 3 if A is the cube:
{A(II), G(II)} = {5832, 1944} and G'(II) = 4491
And of the 2 or 3-digit Fibonaccis there is only one fit:
{A(I), G(I)} = {987, 329} and G'(I) = 923
So, the value G = A / 3 cannot belong to grid III.
 
Now looking at the down clue:
 
a = G – A : G' : Palindrome
G – A cannot belong with G = A / 3 where a = G' or a = Palindrome.
a(II) = ???2, which doesn’t fit with G'(II) = 4491 so G = A / 3 belongs with a = G' in grid I.
Also G(II) can only be Prime (not b / A)and a(II) cannot equal G – A (a Prime minus 5832 cannot equal ???2).
 
We now know:
 
A(I) = Fibonacci = 987
A(II) = Cube = 5832
a(I) = G' = 923
a(II) = Palindrome = 2??2
G(I) = A / 3 = 923
G(II) = Prime
b(II) = 2E = 54
d(I) = B + g
d(II) = A / b = 108
E(II) = Cube = 27
 
And grids I and II look like this:
 
 
Continuing, there are no Fibonacci numbers of the form 7???3 so b(I) = AB and B(I) (Prime or a Square) is in the range 71 to 81.
{A(I), B(I), b(I)} = {987, 71, 70077} {987, 73, 72051} {987, 79, 77973} {987, 81, 79947}
d(I) = B(I) + g(I) = 79 + g(I) = 9? so g(I) is in the range 11 to 19.
g = Fibonacci : F / 2 : Square
{g(I), d(I), F(I)} = {13, 92, 30-39} {12, 91, 24} {16, 95, 60-69}
 
F = 2B : A + e' : Factor of D
F(I) can only be a Factor of D and F(II) = 2B
F(II) = 2B = 882
B(II) = 441 (a Square)
a(II) = 2442
g(II) = 81 or 89
 
f = Prime : 2 × Square : Palindrome
Ending in a 9, f(I) cannot be 2 × Square so must be Prime or a Palindrome
 
At this stage I am getting a bit bogged down so will have a look at placing the six horizontal bars in grid III. Two bars cannot go into the centre column or there will be an odd number of unchecked cells, which must be six. Therefore, because no two grids have bars in the same position, one bar must go in columns 1 and 5 and two bars in columns 2 and 4. There are two possibilities:
 


But the grid on the left has only five down entries leaving the grid on the right as the correct construction for grid III.
 
Continuing with the Across clue:
 
D = EG : g(A – f) : Palindrome
Only D(I) fits g(A-f) = ?9?
g(I) = 12, 13 or 16
A(I) = 987
f(I) = Prime or Palindrome from 909 to 969 = 909, 919, 929, 939, 949, 959 or 969
{g(I), f(I), D(I)} = {12, 929, 696} or {13, 949, 494} are the only fits.
 
But E(I) = Cube + g or Cube – g = ?7?? (Cube = 1728 or 2744)
and only {g(I), f(I), D(I), E(I)} = {12, 929, 696, 2732} fits.
F(I) = 24
d(I) = 91
e(I) = Square or Cube = ?624 = 4624 (68 Squared)
 
Only c(II) fits d' – B = 360
D(II) cannot be a palindrome so must equal EG.
 
C = Prime : E + E' + e : Prime × F
C(II) only fits being Prime and E + E' + e is too large for C(I) (247?) so C(I) = Prime × F
C(I) = 2472 (103 × 24)
c(I) = 7263 = Prime × Cube (269 × 27)
 
This completes grid I:
 
 

Completing grid II was then plain sailing starting with D(II) = EG = 4?00?.
 
The final entry here, e(II), is interesting since it can be either a Prime or Cube of the form 172? or 772? (C(II) is Prime = 61 or 67). There is only one Cube that fits: 1728 but four Primes: 1721, 1723, 7723 or 7727 so e(II) must be a Cube (1728) or else there is not a unique solution. I must note whether or not e(III) could also be a Cube or Prime – I rather hope that it can’t.
 
So, all that remain are the Odd Ones Out to complete grid III:
 

ACROSS (III)
A Prime
B Square factor of D
C E + E' + e
D Palindrome
E Cube + g
F A + e'
G b / A
 
DOWN (III)
a G – A
b Fibonacci
c Prime × Square
d 5G
e Prime
f 2 × Square
g Fibonacci

 

Starting with the 3-digit Fibonacci numbers for b (except 987 = A(I)) there is only one fit for:
 
G = b / A = 29 = 377 / 13
 
e might only be the Cube 512 but that doesn’t fit.
 
After about half an hour, c appears in the grid and as a final check is confirmed to be:
 
c = Prime × Square = 15408 = 107 × 144
 
 
Puzzle completed at 5.26pm, about 3hrs working time, which is quick for me.
 
Post Mortem
 
So, a pure logic problem, no more than an overblown Sudoku, therefore possibly hated by many solvers but I loved this and had a fine old time. I was able to complete it using no more than a calculator, David Wells’ book of Curious and Interesting Numbers (for the listed Squares, Cubes and Fibonacci numbers), my list of Primes up to 10000 (courtesy of BBC Basic in the 80’s) and a mere one and a half sides of A4 for notes. For the first time that I can remember, I went through a numerical puzzle without making a single error so there was none of that tedious backtracking from impasses on this occasion.
 
Initially, I had thought that the construction of grid III might give us an insight into how these puzzles are set so was a bit disappointed to find that all the required information was in the preamble – although it was well concealed.
 
It is rare to have more than one but I would be interested to hear of any opening moves that I might have missed and similarly any shortcuts. For the purposes of this blog, I repeated my opening but this time using BBC Basic to check that the four 4-digit Fibonacci numbers and ten 4-digit Cubes (barring 1000 & 8000) gave a unique solution for A, b and d in grid II:
 
10 READ A
20 X=INT(A/1000)
30 Y=X*10
40 FOR b=Y TO Y+9
50   d=A/b
60   IF d=INT(d) AND d>99 AND d<1000 THEN PRINT A,b,d
70 NEXT b
80 GOTO 10
90 DATA 1597, 2584, 4181, 6765, 1331, 1728, 2197, 2744, 3375, 4096, 4913, 5832, 6859, 9261
 
This gives four solutions but only one fit for b (Fibonacci or equal 2E):
 
A = 5832
b = 54
d = 108
 
I have opted out of the debate on whether or not there is any place for numerical puzzles under the heading The Listener Crossword since I didn’t think that it was going anywhere but, from the mid eighties, they have always been an essential element for me so long may they continue.

2 Responses to “3984 – Odd One Out by Oyler”

  1. Re: Odd One Out by Oyler (Listener number 3984, May 31 2008. I know this is old, but I have only recently joined the Times crossword club, and I’ve been trying out some of the old puzzles.This is only the second maths puzzle I have tried, but I think there is an error in the puzzle, and I am at a loss to understand why both your working shown on the website (https://listenwithothers.com/2008/06/20/listener-3984-odd-one-out-by-oyler/) and Oyler’s own solution arrive at the same (wrong?) answer. I refer to the third puzzle, where F = A + e’. Oyler states in the preamble that “A’ denotes the reverse of A”. In this puzzle, A = 16 and e = 557, so e’= 755. But 755 + 16 = 771, so F should be 771, yet both you and Oyler have F = 768. I assume the winners of the competition also had 768, but I am at a loss as to how this figure was arrived at. I would be grateful if you could illuminate matters for me. Best wishes, Linda Roberts

  2. erwinch said

    Hello Linda,

    I am afraid that you have misread the value for A. Upper case refers to across entries so A=13 in the third grid (a=16).

    Anyway, I hope that you enjoyed the puzzle.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.

 
%d bloggers like this: