## 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 **t2**^{4}. The minimum value of **B7** + **U5** + **s4** is 1061000 + 10000 + 1000 = 1072000, and the maximum is 9861898 + 99998 + 9888 = 9971784, ie **t2**^{4} = 1072000 to 9971784 (remember that the shaded squares must hold an even digit, meaning that the maximum possible value is 8). Thus **t**2 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 = **F2**^{2}.64, fixing **F2**^{2} 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 88**q3** + 2368 = 1536(10**d** + 4), or 11**q3** = 1920**d** + 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 100**x** + **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(100**x** + **y**)**y** = 92162 / 4 = 21233664,

ie (100**x** + **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 + 10**k** 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 = 1000**a** + 100**b** + 10**c** + 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: (100**b** + 10**c** + 6) + □4 = 550,

ie (100**b** + 10**c**) + □4 = 544,

ie (10**b** + **c**) + □ = 54,

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

ie 100**a** + 10**b** + 13**c** = 457

From 2) **c** must be 9 for the LHS to end in 7. This leads to 10**a** + **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.**

## Larry Baum said

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)?

## Richard3435 said

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