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 The bottom line (Posted on 2013-08-05)
```+-----+-----+-----+-----+-----+-----+-----+
| 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 |
+-----+-----+-----+-----+-----+-----+-----+
|  ?  |  ?  |  ?  |  ?  |  ?  |  ?  |  ?  |
+-----+-----+-----+-----+-----+-----+-----+
```
1. Each of the letters represents a different digit from 0 to 9. No number can contain any leading zero.
2. J=0, G=E+A, B=C+I and, D=B-G
3. 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.

 No Solution Yet Submitted by K Sengupta No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 Solution Comment 2 of 2 |

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

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