# Listen With Others

## Listener 4114: Three-Square by Elap

Posted by erwinch on 17 December 2010

Elap has provided us with a numerical puzzle every year since 2003 – let us hope that this is a tradition that will long continue.  I first had to do a bit of revision for I had forgotten the meaning of the term relatively prime, if indeed I ever knew it.  I rather fancied to pronounce primitive as prime-itive in this context but am reliably informed that anyone doing so would be open to ridicule.  Of the four sets of clues, the Threes reminded me of a notable sequence.  Which of the following is the odd one out?

31
331
3331
33331
333331
3333331
33333331
333333331

The last is the only one that isn’t prime (= 17 × 19607843) and there are no more primes in the sequence until 333333333333333331 at 18 digits.

It was the Threes which provided the opening here for me since P2 and j2 intersected (the p’s and q’s in italics are the values derived from the Pythagoras theorem):

333333 = E3 * P2 * j2
Factorising 333333 = 3 × 3 × 7 × 11 × 13 × 37
So {E3, P2, j2} = {117, 77, 37}, {143, 37, 63}, {693, 37, 13} or {777, 33, 13}
But 333 = B3 + N2 + j3 so j2 = 13

Looking at the second Right-angled clue:

E3, E3 – M3, V2 * X2 + Q3
All possible values of p and q for E3 = 693 or 777 (the odd term so p^2 – q^2) were looked at.  So for E3 = 693 (Sq rt > 26) I started with p = 27 and then 28 looking for integer values of q until M3 in 2pq was too big.  27^2 – 6^2 = 693 and 2pq = 324 but M3 = 369 whereas we knew that the final digit was 3.  I found just one fit with M3 = ??3: p = 29, q = 8: E3 = 777, M3 = 313, P2 = 33, N2 = 13, V2 * X2 + Q3 = 905, B3 = 18? or 19?

From 3333333 = P2 * X6 + W3 we find that X6 falls within the range 100980 to 101007 so X2 = 10
Since V2 * X2 + Q3 = 905 then Q3 = ??5, R2 = ?5, S3 = 5?? and g4 =???5
From X3 * X3 = R2 + g4, X3 = 100, X4 = 1009 and g4 = 99?5
From h2 * h2 = S3 + i2 we see that h2 falls within the range 23 to 26, e3 = 12? and D2 = 27
And from e3 * e3 = F5 + Q3, where F5 = ????1 and Q3 = ??5, we see that e3 = 124 or 126

Now looking at the first Right-angled clue:

D2, s3, q3
We know that D2 = 27 (p > 5) so s3 is even = 3?? and q3 = 3??
Just one fit was found: p = 14, q = 13: s3 = 364, q3 = 365, {W3, X6} = {894, 100983} or {564, 100993}

Then the third Right-angled clue:

G2, Q2, k2
G2 = 4? or 6? and k2 = ?3
Since we had k2 = ?3 (p^2 + q^2) the possible values of p and q were limited and we soon had two fits: {G2, Q2, k2} = {45, 28, 53} or {48, 55, 73} and e3 = 124
If G2 = 45 then e3 * e3 – Q3 (285) = F5 (15091) but F5 would equal ?45 ?1 so G2 = 48, Q2 = 55, k2 = 73, F5 = 14821, Q3 = 555, V2 = 35, R2 = 55, g4 = 9945  [Up until this point I had thought that Elap had kindly presented the first two terms in Right-angled clues in the order that could be written p^2 – q^2 then 2pq]

The grid now looked like this:

And so it went: from W2 * W2 = b4 + u2 – we knew that W2 = 89 or 56 and b4 = ?878 or ?978, where 56^2 (3136) could not fit, so W2 = 89, W3 = 894, X6 = 100983, b4 = 7878, u2 =43 and B3 = 18?  But I shall not go any further with the working here.  A full solution should be available on the Listener site but I doubt that it will tally with my route much beyond the first few steps if at all.  I also suspect that it will expose my route as being clumsy and inept.

Of the 22 clues, all but the second Scalene (16th solved) were left until the very end.  The final clue solved was the third Scalene:

G3 + L2, N3 + z2, y3, b4

7878 * 7878 = (504 + x/2 + y/2) * (x/2 + y/2 – 1) * (301 – x/2 + y/2) * (204 + x/2 – y/2) where x was the second digit of z2 and y the third of y3.  I started off using a calculator with x = 0 and y = 0 to 8 step 2 (both would be odd or both even) but got fed up with the calculation and so used BBC Basic.  Not for the first time with the numerical puzzles, I should have stuck with the calculator since x = 1 and y = 5.

So, all the clues were solved and we had five complete rows and five complete columns – what was the common link between these large numbers?  They couldn’t be relatively prime since more than one was even so I first thought that they might be perfect cubes since that would link the threes, squares and three sides of the triangles – to my mind at least – but no.  I then tried triangular numbers and this fitted.

The general formula for the nth triangular number, ½n(n+1), was useful in completing the grid, the positive root of the quadratic equation telling us roughly where to look, but it was troublesome in the two cases where we did not know the first digit to have to repeat the calculations – I started with 9 as the first digit and went down.  Here is my final grid:

As ever with Elap, this was all tremendous fun and I eagerly await his next.  However, of the four sets of clues, I couldn’t see what was thematic about the Squares other than the link to the title and possibly Pythagoras.  Here is a final thought: if you wanted to play snooker with 96972777028 red balls, the 440392nd and largest triangular number in the grid, then you would need a frame with sides over 23km long!