Listen With Others

Listener 4216, Evens: A Full Solution by Elap

Posted by Listen With Others on 8 December 2012

In these notes, □ denotes a digit and a full stop denotes multiplication (but in a few places, x has been used where that is clearer). ‘Side’ implies the length of the side.

The puzzle has been designed to be enjoyably solved by deduction, not by the defeatist, sledgehammer approach of a program or spreadsheet.

First of all, let’s highlight the cells which are known to contain an even digit – the preamble has stated or implied that all the clued numbers are even:

Side Perimeter Area
Square 1: C3 + G3 + V2 D2.H2 + J4 A6 + R6 + d4
Square 2: D2 + h4 n4 + p4 S4.h4 + S5
Square 3: L3 m3 Q4 + R4 + c4
Square 4: N2.q3 + n3 b2.d4 d4.k3.r2
Square 5: T2 + W4 + a2 i5 + j2 P8 + a8 + u7
Square 6: T2 + t2 M2 + u3 F2.F2.j2
Square 7: a2.a2 a2.j2 D4 + e2 + j5
Square 8: e2 + g3 M2 + N4 E3 + K6 + f4
Square 9: t2.t2 P3.t2 B7 + U5 + s4

Square 7: a2.a2, a2.j2, D4 + e2 + j5

Since the perimeter is four times the length of a side, j2 = 4.a2, and this means that a2 must be less than 25. Since, from Square 6’s area, j2 must be a square, a2 = 16 and j2 = 64. This leads to D4 + e2 + j5 = 65536, or D4 + e2 + 64□□□ = 65536. D4 therefore begins with 1:

Square 9: t2.t2, P3.t2, B7 + U5 + s4

P3 = 4.t2 = □□6, and so t2 = <□4 or □9. However, t2 must be even, and so t2 = □4. The area is t24. The minimum value of B7 + U5 + s4 is 1061000 + 10000 + 1000 = 1072000, and the maximum is 9861898 + 99998 + 9888 = 9971784, ie t24 = 1072000 to 9971784 (remember that the shaded squares must hold an even digit, meaning that the maximum possible value is 8). Thus t2 is in the range 34 to 56. Since we know that t2 ends in 4, the only possible values are 34, 44 and 54, corresponding to P3 = 136, 176 and 216. However, the first digit of P3 is even. Therefore, t2 = 54 and P3 = 216:

Square 5: T2 + W4 + a2, i5 + j2, P8 + a8 + u7

For the area we have 216□□□□□ + 16□□1□54 + 416□□1□. The minimum value of this is 21600000 + 16001054 + 4160010 = 41761064, and the maximum is 21698998 + 16991854 + 4169918 = 42860770, fixing the side in the range 6464 to 6546. This is T2 + W4 + a2, ie □□ + □□□4 + 16 = 6464 to 6546. Since the second digit of W4 is even, and t2 provides 4 as the last digit, we have W4 = 64□4.

Square 3: L3, m3, Q4 + R4 + c4

Since m3 is now known to end in 4, L3 must end in 6 (from m3 = 4.L3) – it can’t end in 1 because it must be even. From the size of the area (less than 30000), L3 must begin with 1.

Square 5: T2 + W4 + a2, i5 + j2, P8 + a8 + u7 again:

The minimum value of the side is 10 + 6404 + 16 = 6430 and the maximum is 98 + 6484 + 16 = 6598. The perimeter is therefore in the range 25720 to 26392. The second digit of i5 is therefore 5 or 6:

Square 2: D2 + h4, n4 + p4, S4.h4 + S5

The side is 1□ + □□□4 and the perimeter is 56□□□ + 4□□□ = 9000 to 11986 (n4 and p4 are both even, as is the third digit of n4). The side is therefore 2250 to 2998, and so h4 begins with 2.

Square 8: e2 + g3, M2 + N4, E3 + K6 + f4

The side is □□ + □□6 and the perimeter is □2 + 2216. The perimeter is therefore in the range 12 + 2216 to 92 + 2216, ie 2228 to 2308. This makes the side 557 to 577. The area is therefore 310249 to 332929. This means that K6 starts with 3. e2 + g3 is therefore □□ + □36 = 557 to 577. Therefore, g3 = 536.

Square 6: T2 + t2, M2 + u3, F2.F2.j2

The perimeter is □2 + 416, ie it’s in the range 428 to 508, making the side 108 to 126. The area is therefore 11664 to 15876 = F22.64, fixing F22 in the range 183 to 248. Since we know that F2 is even, we have F2 = 14. The side is therefore 112, giving T2 = 58, and the perimeter is 448, giving M2 = 32.

Square 5: T2 + W4 + a2, i5 + j2, P8 + a8 + u7 again:

The area is 216□□□32 + 16□□1854 + 416□□18, and this ends in 04. The side is 58 + 64□4 + 16, ie 64□4 + 74, which ends in 8, and the square of this value therefore ends in 04. The square root of a number ending in 04 must end in 02, 48, 52 or 98. W4 is therefore 6424 or 6474. However, we know that the third digit is even form n3, and so W4 = 6424. The perimeter is therefore 4 x (58 + 6424 + 16) = 25992, giving i5 = 25928:

Square 2: D2 + h4, n4 + p4, S4.h4 + S5 again:

The area is □□58 x 2□□4 + □□584 and so it ends in 6. D2 + h4 therefore ends in 4 or 6, meaning that D2 must end in 0 or 2. Since e2 cannot start with a zero, D2 must end in 2, and so we have D2 = 12.

Square 9: t2.t2, P3.t2, B7 + U5 + s4 again:

We have

 B7 □□61214 +U5 □□□58 +s4 928□ ────── 8503056

From this, B7 = 8461214, s4 = 9284, and we can deduce that U5 = 32558.

Square 2: D2 + h4, n4 + p4, S4.h4 + S5 again:

The side is 12 + 25□4, the perimeter is 5928 + 4□□□, and the area is 2558 x 25□4 + 25584. We therefore have (12 + 25□4)2 = 2558 x 25□4 + 25584. This gives h4 = 2544, producing a perimeter of 10224, and so p4 = 4296. This gives j5 = 64296, and from Square 7, e2 = 26.

Square 3: L3, m3, Q4 + R4 + c4 again:

h4 has supplied m3 = 544. This gives L3 = 136.

Square 8: e2 + g3, M2 + N4, E3 + K6 + f4

We have a side of 562, producing an area of 315844:

 K6 3136□□ +f4 1□3□ +E3 214 ─────── 315844

Since f4 can’t end with 0 (otherwise Q4 would begin with 0), there must be a carry-over from the units column, and so the fifth digit of K6 must be 9. The second digit of f4 must then also be 9:

 K6 31369□ +f4 193□ +E3 214 ─────── 315844

Square 3: L3, m3, Q4 + R4 + c4 again:

Since L3 is 136, the area is 18496. We therefore have □□32 + □322 + 4□□2 = 18496, and so c4 is 4□42.

Square 4: N2.q3 + n3, b2.d4, d4.k3.r2

Letting the first digit of b2 be d, we have 4(22.q3 + 592) = □4 x 1536, ie 88q3 + 2368 = 1536(10d + 4), or 11q3 = 1920d + 472. We know that d is even (since A6 is referenced), and so the only valid values are 2, 4, 6 or 8. d = 2 gives q3 = 392 (there is no point in trying the other values of d because adding small multiples of 3840 won’t create another number which is divisible by 11), resulting in a side of 9216. This gives b2 = 24. Do you see that it was not necessary to use a program or spreadsheet to get this result?

For the area, we have 1536.k3.r2 = 92162, and we know that the first digit of k3 is even, and that k3 and r2 are even, and that r2 is the same as the last two digits of k3. Let half of k3 be 100x + y, where x is 1, 2, 3 or 4 (representing half the first digit of k3) and y is in the range 1 to 49 (representing half of r2). Then we have

1536(100x + y)y = 92162 / 4 = 21233664,

ie (100x + y)y = 13824 where x = 1, 2, 3 or 4.

Is this the first time since our schooldays that we’ll need to solve some quadratic equations?

Let’s express 13824 (= 29. 33 ) as the product of two numbers which have a difference of a few hundred:

 12 x 1152 Difference too large 16 x 864 18 x 768 24 x 576 27 x 512 32 x 432 This could be it – a difference of 400 36 x 384 48 x 288 54 x 256 64 x 216 72 x 192 96 x 144 Difference too small

Thus we have x = 4 and y = 32. Since half of k3 was 100x + y, we have k3 = 864 and r2 = 64 – and we didn’t need to solve any quadratic equations after all!

Square 8: e2 + g3, M2 + N4, E3 + K6 + f4 again:

f4 is now known to be 1938, leading to K6 = 313692.

Square 3: L3, m3, Q4 + R4 + c4 again:

We now have the first digit of Q4, and so the area (18496) = 8□32 + □322 + 4□42. Since the second digit of Q4 is the same as the first digit of R4, the only possibility is 8532 + 5322 + 4642, and so we have c4 = 4642, R4 = 5322 and Q4 = 8532.

Square 5: T2 + W4 + a2, i5 + j2, P8 + a8 + u7 again:

For the area we now have 216□8532 + 16□□1854 + 416□□18 = 64982 = 42224004:

 P8 216□8532 +a8 16□□1854 +u7 416□□18 ────── 42224004

It is easy from here to deduce that u7 = 4163618 and a8 = 16361854 (remember that u7 and a8 overlap). This gives P8 = 21698532.

Square 1: C3 + G3 + V2, D2.H2 + J4, A6 + R6 + d4

The side is 612 + □□6 + □4; the perimeter is 12.□6 + □□□6, and the area is 816□□2 + 532216 + 1536 – this must be a perfect square. We therefore have 1349754 + 10k is a square, where k is the two missing digits (ie in the range 0 to 99). The square root of 1349754 must be only a little less than the side, and is 1161.789, and so the side must be 1162. This gives an area of 1350244, making k = 49. This gives A6 = 816492.

Considering the side, we have 612 + □□6 + □4 = 1162, ie □□6 + □4 = 550, and for the perimeter we have 12.□6 + □□□6 = 4648. Let J4 = 1000a + 100b + 10c + 6, where a, b and c are 1 to 9 (none of them can be zero because each digit starts a clued number). Then we have

1) For the side: (100b + 10c + 6) + □4 = 550,

ie (100b + 10c) + □4 = 544,

ie (10b + c) + □ = 54,

2) For the perimeter: 12.(10c + 6) + (1000a + 100b + 10c + 6) = 4648,

ie 100a + 10b + 13c = 457

From 2) c must be 9 for the LHS to end in 7. This leads to 10a + b = 34, and so a = 3 and b = 4, in turn leading to J4 = 3496 and V2 = 54.

The clues have now been solved, but six cells are still empty:

We now have to look for the common property. Since we have been told that the sides of the squares share this property, let’s list them:

 Square 1: 1162 Square 2: 2556 Square 3: 136 Square 4: 9216 (Perfect square) Square 5: 6498 Square 6: 112 Square 7: 256 (Perfect square) Square 8: 562 Square 9: 2916 (Perfect square)

Some of these numbers are square – but only three of them.

What is going on?

• The theme of the puzzle is squares
• A feature is that the grid entries sometimes wrap round to the beginning of the row or column.
• There are no zeroes or sevens in the grid
• The preamble: “…The final grid should show what would otherwise be missing from it”

Ah ha! The side of each square is either a perfect square or a rotated version of one – and there aren’t any zeroes or sevens there, either!

 112 121 rotated 136 361 rotated 256 Already square, and 625 rotated 562 256 or 625 rotated 1162 2116 rotated 2556 5625 rotated 2916 Already square 6498 8649 rotated 3 digits to the right 9216 Already square

Each row and column of the grid is also a square or the rotation of a square.

Let’s start with the second row. The square could end with 84, 61, 21, or 4□, and so the possibilities are 8461214□, 61214□84, 214□8461 or 4□846121. Only the last one works: 48846121 = 69892. Now the first two columns can be tackled. By approaching row 7 next, and then the last two columns, we arrive at this (the bars indicate where the perfect squares begin, but were not required to be entered):

So there are some zeroes in the grid after all!

The preamble said: “After solving the clues, the grid must be completed so that every row and column shares a thematic property, given the right starting point. One of them is doubly thematic; solvers are to draw a straight line through this number’s digits.”

We’ve found the common property, but one of them is “doubly so”.

Investigation reveals that the first row is the rotation of two different squares – 81649296 = 90362 and 92968164 = 96422, and so we can draw a line through the digits:

The preamble also said

“A seventeenth number in the grid which has the same property as the rows and columns must also have a line drawn through its digits, so that the final grid then shows what would otherwise be missing from the puzzle.”

We therefore need to find another rotated square in the grid. But what is missing from the puzzle? Seven. There are no 7s in the grid (there weren’t any zeroes either, but a couple have appeared now).

For the final grid to show a 7, the top right / bottom left diagonal should have a line through it, and further investigation reveals that 12645136 (which reads in the direction bottom left / top right) is 35562.

We can therefore draw a line through these digits to complete the puzzle:

The double-rotation feature of the first row was echoed in the sides for Squares 7 and 8.

The puzzle’s title was a rotation of “seven”.

Elap.

1. Larry Baumsaid

All very nice and clear until I got to:

“The square could end with 84, 61, 21, or 4□, and so the possibilities are 8461214□, 61214□84, 214□8461 or 4□846121”

Why not 1214×846 or x8461214 (I realize neither of those have a solution, but why aren’t they candidates)?

2. Richard3435said

Any even square is a multiple of 4. A number ending 14 or 46 is even, but not a multiple of 4.

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