# Listen With Others

## Listener 4125: Elementary Number Theory by Oyler

Posted by erwinch on 11 March 2011

So, the first numerical Listener of the year and now a week earlier than previously.  This is Oyler’s tenth Listener and he certainly needs no introduction from me.  I am not going to trawl back through my files but I do not recall seeing clues of this form before.  We were told which of 18 statements (A to R) applied to each entry so the statements that did not apply were of equal use to the solver.  There was one statement (R: It has a repeated digit) that did not apply to any of the clued entries.  The expectation was that it would apply to the two unclued entries: 23ac and 5dn, both of 9 digits.

There appeared to be numerous starting points but I started with 20ac to which only one statement applied: I: It is a perfect square (3).  There are 22 3-digit perfect squares and all but one could be eliminated by considering the 17 statements that did not apply – statement R was particularly useful here.  20ac = 196

On to 20dn HK (1?) where we told that the sum of the first and last digits were a perfect square (H), so 20dn = 10, 13 or 18, and that it is the sum of four consecutive integers (K) The general formula for the sum of four consecutive integers is 4n + 6 where n is the first integer therefore they are all even numbers (from 10 step 4): 20dn = 10 or 18

32dn A B E H M P (3) had the greatest number of statements that applied:
A: It is a prime number.
B: Its digit product is odd – all digits are odd.
E: It has a remainder of 1 when divided by 4 – the number ends with 1, 3, 5, 7 or 9 but not 5 (A) nor 1 (P) here.
H: The sum of its first and last digits is a perfect square (ie 1, 4, 9 or 16 but not 1 or 4 here).
M: Its digit sum is a prime number (ie 11, 13, 17, 19 or 23 here).
P: Its digit product is a three digit number (all digits are >2 here).
Again R was very useful since we knew that it had no repeat digits: 32dn = 937

39dn IJ (2): it is a perfect square (I) and its digits are all even (J): 39dn = 64

38dn BCFG (2) and all digits are odd (B):
C: It is the product of two distinct primes that are less than 100.
F: It is a multiple of 5.  It equals 5 × 3, 7 or 19.
G: It is a triangular number.  38dn = 15

32ac BCDEL = 9?  We only needed to consider one statement here:
L: Its first digit is 3 times the units digit.  32ac = 93
D: It has a remainder of 2 when divided by 7.

37ac HQ = ?31  We only needed to consider H since R did not apply: 37ac = 831
Q: It is a composite number with a three-digit prime factor.  831 = 3 × 277

41ac CEH = ??7  From H it equals 2?7 or 9?7 and from E it equals 217, 237, 257, 297, 917, 937 or 957.  But it is not prime (A) and not all digits are odd (B), which leaves 217, 237 or 297.  Also, its digit product is not a three-digit number (P) leaving 217 or 237.  From C: 217 = 7 × 31 and 237 = 3 × 79 so 41ac = 217 or 237

42ac EP = 54?  From P and R it is > 545 and from E it equals 549: 42ac = 549

26ac EFMPQ (3)  From F and Q it equals ??5 and from M and P, the first two digits are both even and greater than 2.  Possibilities are 465, 645, 485, 845, 685 or 865.  These all comply with E but not M or Q: 26ac = 685 or 865

36ac EFIM (2)  36ac = 25 is the only possible fit.

14ac AB (2) and 1dn CF (3)  1dn must end with 5 so 14ac = 53 or 59.  15dn CL (3) = 3?1 or 9?3 but H does not apply so 15dn = 9?3 and 14ac = 59

18ac CEGM (2)  There are ten two-digit triangular numbers (G) but only 21 and 45 comply with E while F does not apply so 18ac = 21

9dn EHLQ = ??1  From L and H it equals 3?1 and from Q it equals 3 × 107 or 3 × 127: 9dn = 321 or 381

13ac KM = ?2 or ?8  From K it equals 22, 42, 62, 82, 18, 38, 58, 78 or 98.  But M excludes all but 38, 58 or 98: 9dn = 381  But 38 = 2 × 19 and 58 = 2 × 29 yet C does not apply so 13ac = 98

21ac DFPQ (3) must end with 5 so 22dn HNOP = 5?4 to comply with H.  O: It is a multiple of 12 so 22dn = 564 (504 does not comply with P).  N: It has at least 12 factors (including itself and 1)

39ac AM = 61 or 67 but 34dn P = ?19 or ?79 but must equal ?79 to comply with P: 39ac = 67  33ac CJK (2) is even (J) so 34dn = 279, 479, 679 or 879 but A, C and Q do not apply (479 is prime, 679 = 7 × 97 and 879 = 3 × 293) so 34dn = 279

The grid now looked like this with one column complete:

We could see that the completed column 9 contained the numbers 1 to 9 once only and that there were no repeated numbers in any row, column or major diagonal so far.  As a rule it is futile to make any assumptions with these puzzles but might this be the property that the completed grid was to display?  There was a 9 in column 2 so would 94 fit for 35ac CKM?  94 = 2 × 47 = 22 + 23 + 24 + 25 and 9 + 4 = 13 so indeed it did while it did not comply with any remaining statement.  However, I was very wary so continued by placing further entries in the top right corner of cells.  I had a moment of doubt when two sixes appeared on the main SW/NE diagonal but the grid was soon complete with all rows and columns complying with the assumption.

Factorising the two unclued entries:

23ac 139854276 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 73 × 73

The statements that apply are DINO

5dn 627953481 = 3 × 3 × 8353 × 8353

The statements that apply are EI

DINO + EI is an anagram of Iodine, the word to be written below the grid.

So, the puzzle was complete.  However, there was a little extra hidden in the grid.  Ninas are rare in the Listener since the grid is usually milked for all its thematic material.  It is bad enough trying to find things in the grid even when you are looking for them but I think that the last we saw was Pacific Ocean appearing unnoticed by most solvers, including myself, in News Quiz by Stan back in November.  A clueing device seen a couple of times over the past year is, for example, at No.10 to denote NEON or NE, well, at No.53 we have Iodine.  53 appears seven times in the grid, including diagonals and reversals, but only one is of note and the cells should really have been required highlighting – in an appropriate shade of course:

This also gives us the elemental symbol, albeit in a somewhat squat font, and to argue that highlighting any other 53 could also be judged correct is absurd – our highlighting is central and is found in the two entries that supplied the Iodine.

To sum up, this did rather live up to its title in being elementary, in all senses, but I found it a most enjoyable and absorbing puzzle, certainly DEF H! (see puzzle number)

### 4 Responses to “Listener 4125: Elementary Number Theory by Oyler”

1. ### John Poolersaid

7 of the solutions were IDONEI numbers. It seemed to me that IDONEI was a better thematic word than IODINE which is rather far fetched.

As for the solution of the crossword. Put the statements into a QBASIC program, eg for statement D,(N – 2)/7 = INT((N – 2)/7). Many of the clues are solved immediately and the others follow easily after.

2. ### erwinchsaid

Idonei or numeri idonei do not appear in Chambers which probably precludes this as a theme.  It should perhaps be remembered that thematic numerical Listeners were all but unknown ten years ago – the theme was simply that they were numeric and they were much appreciated as such by at least two-thirds of solvers (see here).  I thought that the play on element made Iodine a perfectly acceptable theme and anyway we should not expect too much.

Using computers to crack these puzzles is rather missing the point that they are entertainments to be savoured by the solver and not a chore to be dispatched as soon as possible.  All well and good if you find using the computer entertaining but I would suggest that your time might be better spent elsewhere.

3. ### John Poolersaid

Thanks Erwinch. The puzzle did not actually say that the thematic word had to be in Chambers. Euler’s numeri ideoni could easily have been a statement so seemed particularly relevant.

The fun was in writing a program to crack the puzzle. Some of the statements were not so obvious,
eg triangular numbers have
N = (INT(SQR(2*N))*(INT(SQR(2*N)) + 1)/2
and the statement for a three digit prime factor requires several lines of program.

You did not mention that the solution also had nine 3X3 squares each of which had the nine numbers like a suduko.

4. ### erwinchsaid

Thanks John, I honestly had not realised that it was a full 9 × 9 Sudoku until I read the other blogs and comments – I assumed that it was a Latin square and thought that the main diagonals might also comply.  Looking at it now, it is an impressive construction from Oyler.  I did try Sudoku when it first appeared in the UK (early 2000’s?) but found the logical process too samey for my taste and so have not solved one since.

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