Listener 4216: Evens by Elap
Posted by Dave Hennings on 7 Dec 2012
Oh dear! This week it’s Elap, he of triangular numbers from two years ago. Here we had a wide open spaced 8×8 grid only populated with clue letters, no bars. Unlike Threesquared, which dealt with triangles, this one had clues to the sides, perimeters and area of squares. Eventually, every row and column would need to be completed thematically, and a couple of lines drawn. So at the bottom of today’s effort is my attempt at a detailed solution.
It was, I have to say, fairly obvious that the two lines which would end up showing what was otherwise missing from the puzzle would be in the shape of the number 7. That was made ever so obvious by the title which was Seven with S moved to the end. However, as with many early penny drops, it didn’t help much.
I started by highlighting those squares which contained the last digit of a clue term, eg the last digit occupied by C3, G3 and V2 in 1ac. Square 6 was the place I started, and two entries went in the grid fairly early on … well, after about 40 minutes! As with many mathematicals, it was like chipping away at a pyramid — although you felt you were progressing, there seemed very little to show for it. But I got there in the end.
After all the clues were solved, the grid was complete … except for those pesky six empty squares, three in the top left and three in the bottom right. Unlike the triangular number in Elap’s last puzzle, it didn’t take too long to realise and verify that all the rows and columns formed squares if started in the right place. Moreover, the main SENW diagonal was a square, not to mention 1ac twice: 81649296 and 92968164.
It has to be said that refining, tarting up and keying in my original notes took well over twice the time that it took me to solve Elap’s puzzle in the first place, which was about 4½ hours. I hope you’ll forgive this first attempt, which is nothing compared to erwinch’s efforts, especially in their lack of colour. I’ll try and do better next time!
In this table, I have marked unknown digits with the • character, and even digits with an E, eg 12•E•.
No  Clue Details  Logic  Extra Info  

1  Sq 6  A=F2.F2.j2  ∴ j2 is a square  j2=64 
2  Sq 7  S=a2.a2 P=a2.j2 
j2=4.a2  a2=16 
3  Sq 7  from 2  D4+e2+j5=65536  j5=64••• ∴ D4=1••• 
4  Sq 2  S=D2+h4 P=n4+p4 
p4=4••• ∴ max(n4+p4)=9999+4999=14998 ∴ max(D2+h4)=14999 with 4=3749% but h4 begins even 
h4=2•EE 
5  Sq 9  S=t2.t2 P=P3.t2 A=B7+U5+s4 
P3=4t2 ∴ max(P3)=4.99=396 ∴ P3=1•• or 2•• or 3•• but P3=E•6 divisible by 4 ∴ P3=216 or 236 or 256 or 276 or 296 ∴ t2= 54 or 59 or 64 or 69 or 74 but t2=•E ∴ t2= 54 or 64 or 74 ∴ P3=216 or 256 or 296 but max(B7+U5+s4)=9999999+99999+9999 ∴ max(t2)=56 
t2=54 P3=216 u3=416 N4=•216 
6  Sq 8  S=e2+g3 P=M2+N4 A=E3+K6+f4 g3=••6 
M2+N4=4(e2+g3) M2+N4=•2+22•6 =(12+2206)→(92+2296) =2218→2388 ∴ e2+g3=550*→598* ∴ g3=451→588 (e2+g3)^2=E3+K6+f4 
g3=536 d4=1536 
7  Sq 4  S=N2.q3+n3 P=b2.d4 A=d4.k3.r2 d4=1536 
(b2.d4/4)^2=d4.k3.r2 b2^2.d4^2=16.d4.k3.r2 b2^2.1536=16.k3.r2 b2^2.96=k3.r2 max(k3.r2)=999.99=98901 digits 2&3 k3 = r2 b2=EE trying eg 384.84, 464.64, 466.66, etc 
b=2• 
8  Sq 6  S=T2+t2 P=M2+u3 A=F2.F2.j2 t2=54 
M2+u3=428→508 ∴ F2.F2.j2=(428/4)^2→(508/4)^2 =11449 or 12544 or 13689 or 14884 or 16129 but is divisible by j2=64 ∴ F2.F2=12544/64=196 ∴ F2=14 
F2=14 T2=58 M2=32 b2=24 from 6: from 8 
9  Sq 7  S=a2.a2 P=a2.j2 A=D4+e2+j5 a2=16 
a2.a2=256 D4+e2+j5=65536 =1214+26+j5 
j5=64296 
10  Sq 9  S=t2.t2 A=B7+U5+s4 B7=••61214 
B7+U5+s4=8503056 min(U5)=10058 ∴ B7=8461214 

11  Sq 5  S=T2+W4+a2 P=i5+j2 A=P8+a8+u7 T2=58 
T2+W4+a2=58+•••4+16=•••8 i5+j2=4.•••8=•••••8 ∴ i5=••••8 
i5=25••8 A6=816••2 
12  contd  i5+j2=25008+64→25998+64 =25072→26062 T2+W4+a2=6268→6515 ∴ W4=6194→6441 but W4=•EE4 ∴ W4=6EE4 =6024 or 6044 or 6064 or 6084 or 6204 or 6224 or 6244 or 6264 or 6284 or 6404 or 6424 ∴ i5+j2=24392 or 24472 or 24552 or 24632 or 25112 or 25192 or 25272 or 25352 or 25432 or 25913 or 25992 but digit 4 i5 = digit 3 W4 A=P8+a8+u7=216•8•32+16••1854+416••18 
W4=6424 i5=25928 n4=5928 s4=9284 

13  Sq 9  S=t2.t2 A=B7+U5+s4 t2=54 
A=8503056 =8461214+U5+9284 
U5=32558 
14  Sq 2  S=D2+h4 P=n4+p4 A=S4.h4+S5 n4=5928 
P=10220 ∴ P=2556=12+h4 ∴ h4=2544 

15  Sq 4  S=N2.q3+n3 P=b2.d4 
P=24.1536=36864 ∴ S=9216=22.q3+592 ∴ q3=392 
q3=392 
16  Sq 3  P=m3 S=L3 A=Q4+R4+c4 m3=544 
L3=136
A=18496 by inspection, Q4=8532, R4=5322, c4=4642 
L3=136 
17  Sq 5  P=i5+j2 A=P8+a8+u7 i5=25928 
A=42224004 P8=216•8532+ a8=16••1854+ u7= 416••18 by inspection, P8=21698532, a8=16361854, u7=4163618 

18  Sq 8  S=e2+g3 A=E3+K6+f4 e2=26 
A=315844 E3= 214+ K6=3136••+ f4= 1•38 by inspection, K6=313692, f4=1938 

19  Sq 1  S=C3+G3+V2 P=D2.H2+J4 A=A6+R6+d4 
A=(816002+532216+1536)→ (816992+532216+1536) =1352754→1353744 ∴S=1162 A=1350244 

20  contd  By inspection!! G3=496, H2=96, J4=3496, V2=54 
Leave a Reply