+-----+-----+-----+-----+-----+-----+-----+
| EBF | EHG | EFG | EBE | AJD | EFJ | EIF |
+-----+-----+-----+-----+-----+-----+-----+
| ? | EFA | EIA | AJJ | EHC | EFI | EII |
+-----+-----+-----+-----+-----+-----+-----+
| EHD | EFD | ? | ACJ | EHB | EIH | EIH |
+-----+-----+-----+-----+-----+-----+-----+
| EFE | EIE | EBB | EHJ | ? | EBD | AJH |
+-----+-----+-----+-----+-----+-----+-----+
| EIJ | ? | AJB | EHI | EHG | EBG | EHA |
+-----+-----+-----+-----+-----+-----+-----+
| EIC | EBI | EHH | ? | EBA | AJG | EFC |
+-----+-----+-----+-----+-----+-----+-----+
| ? | ? | ? | ? | ? | ? | ? |
+-----+-----+-----+-----+-----+-----+-----+
- Each of the letters represents a different digit from 0 to 9. No number can contain any leading zero.
- J=0, G=E+A, B=C+I and, D=B-G
- Each of the columns add up to 1990. So does each of the rows and, each of the main diagonals.
Determine the seven 3-digit numbers missing from the bottom line.
Possibility of E is only one i.e. =2
Possibility of A= 3,4,5 but based on other conditions and total of units place of row one A can only be 3.
Hence G= 5 and D=4, B=9 and F=7 from conditions and total of units place of row one. From total of tenth place C=1 since H+I=14.
So H,I =8,6
H=8 and I=6 satisfies the condition of totals of all rows and columns. Hence the grid is.
297 285 275 292 304 270 267
331 273 263 300 281 276 266
284 274 297 310 289 268 268
272 262 299 280 275 294 308
260 272 309 286 285 295 283
261 296 288 276 293 305 271
285 328 259 246 263 282 327
Hope I didnt make any error in typing.
|
Posted by Salil
on 2013-08-08 03:34:21 |