![]() | ANSWERS PART FOUR |
|
4 3 8 7 9 1 2 5 6 |
|
6 3 7 x 1 3 = 8 2 8 1 - + - 4 4 1 x 9 = 3 9 6 9 1 9 6 x 2 2 = 4 3 1 2 |
|
7 4 9 4 5 6 4 4 9 4 3 7 4 5 2 9 9 6 3 4 1 5 4 4 |
|
1 7 1 7 7 1 1 7 1 1 1 9 7 1 1 9 7 1 3 1 8 4 1 |
|
1 3 7 1 9 3 4 1 1 1 2 3 3 1 3 7 2 6 4 4 1 |
Obviously B and D cannot hide two nines.otherwise the line EII would run into four figures.thus B and D must hide, not necessarily in that order, 8 and 7. If D covered 7 and B, 3, line EII would have to run into four figures, in order to have two ones at the end. Terefore B covers 7, and D, 3 and A also 3. Now C must hide either 8 or 9. ottherwise, IFGH would not be a four figure number. But 8 times 137 would mean that F conceals a Zero. which is inadmissible. So C covers a 9, etc.
| The required squares are ABCD, and CBEF. The three cuts are along DA, CB, and DF. SD is 3 cm long. |
![]() |
|
4 3 5 6 2 0 5 1) 8 9 3 2 0 5 6 8 2 0 4 7 3 0 1 6 1 5 3 1 1 4 8 5 1 0 2 5 5 1 2 3 0 6 1 2 3 0 6 |