Danger

[cinelli]

.
   RED
   FOR x
------
****
 ****
   ***
------
DANGER

Cinelli

[Gengulphus]

   RED
   FOR x
------
****
 ****
   ***
------
DANGER

Spoiler...

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First, R cannot be 0 because it's the leading digit of RED. Furthermore, RED * R is a three-digit number, which it won't be if R >= 4, and R also cannot be 1, as the final product DANGER would end with D rather than R if it were. So R must be 2 or 3.

Now move on to D. It is the leading digit of DANGER, so cannot be 0. DANGER is less than RED * 1000 = RED000, so D <= R, and since D and R must be distinct digits, D < R. So D must be 1 or 2. Furthermore, R is the last digit of RED * FOR, which is the last digit of D*R, and neither of the possible values of R allows D = 2. So D = 1.

Now look at E, the last digit but one of DANGER = RED * FOR. It is the last digit of the sum of:

* The tens digit of D*R, the product of the last digits of RED and FOR. Given what we know about D and R, this is 0.

* The product E*R of the last-but-one digit of RED and the last digit of FOR.

* The product D*O of the last digit of RED and the last-but-one digit of FOR.

Since D = 1, that means that E is the last digit of E*R + O. So E*R + O - E is a multiple of 10, which means that O is the digit needed to make E*(R-1) up to a multiple of 10.

We have sixteen possible values for RED, since D is 1, R is 2 or 3, and E is a digit other than D or R. For each of those, we can calculate O from the above and eliminate the value of RED if that causes a digit duplication or causes O to be too small for RED * O to be a four-digit number:

RED = 201: O=0 (eliminated)
RED = 231: O=7
RED = 241: O=6
RED = 251: O=5 (eliminated)
RED = 261: O=4
RED = 271: O=3 (eliminated)
RED = 281: O=2 (eliminated)
RED = 291: O=1 (eliminated)
RED = 301: O=0 (eliminated)
RED = 321: O=6
RED = 341: O=2 (eliminated)
RED = 351: O=0 (eliminated)
RED = 361: O=8
RED = 371: O=6
RED = 381: O=4
RED = 391: O=2 (eliminated)

For each remaining value of RED, we know the last two digits of FOR and know that 123456 <= DANGER <= 198765, so 123456/RED <= FOR <= 198765/RED. This allows us to list the possible values of FOR, eliminating those that duplicate digits:

RED = 231: FOR = F72 and is in the range 535 to 860, so is 572, 672 or 772 (eliminated)
RED = 241: FOR = F62 and is in the range 513 to 824, so is 562, 662 (eliminated) or 762
RED = 261: FOR = F42 and is in the range 474 to 761, so is 542, 642 (eliminated) or 742
RED = 321: FOR = F63 and is in the range 385 to 619, so is 463 or 563
RED = 361: FOR = F83 and is in the range 342 to 550, so is 383 (eliminated) or 483
RED = 371: FOR = F63 and is in the range 333 to 535, so is 363 (eliminated) or 463
RED = 381: FOR = F43 and is in the range 325 to 521, so is 343 (eliminated) or 443 (eliminated)

For each remaining combination of RED and FOR, we can calculate DANGER = RED * FOR and eliminate ones that contain duplicate digits, either within themselves or with the F and O of FOR:

231 * 572 = 132132 (eliminated)
231 * 672 = 155232 (eliminated)
241 * 562 = 135442 (eliminated)
241 * 762 = 183642 (eliminated)
261 * 542 = 141462 (eliminated)
261 * 742 = 193662 (eliminated)
321 * 463 = 148623 (eliminated)
321 * 563 = 180723
361 * 483 = 174363 (eliminated)
371 * 463 = 171773 (eliminated)

That gives the solution - as a final check, do the original long multiplication:

   321
   563 x
------
1605
 1926
   963
------
180723

Gengulphus

[cinelli]

Well solved, Gengulphus. This type of puzzle is best when it has a unique solution, as here.

Cinelli