Odd/even

[cinelli]

This “find the digits” puzzle in this multiplication is a little different. All that is given is whether each digit is odd or even. Nought, if it appears at all, is considered to be even.

. 
  EOEE
   EOE x
------
EOOE
OEOEE
  OOEE
------
EOOOEE

Cinelli

[UncleEbenezer]

This “find the digits” puzzle in this multiplication is a little different. All that is given is whether each digit is odd or even. Nought, if it appears at all, is considered to be even.

. 
  EOEE
   EOE x
------
EOOE
OEOEE
  OOEE
------
EOOOEE

Cinelli

The leftmost E of EOEE and the two Es of EOE must each be 2 or 4, to generate the two 4-digit numbers.
The O of EOEE must be high enough to generate a carry for OOEE but not for EOOE. 1 is too low; 5 too high, so it's a 3 and we have

So that's
23EE
2O4

The remaining O must be high enough to generate 5 digits when multiplied by 23EE, so it's [579].
The rightmost E of 23EE multiplies by 4 with even carry and by 2 with odd carry, so it's [68].

So,
23E[68]
2[579]4

That's just one remaining E. 2*(E+1) must generate odd carry, 4*(E+2) even. Only 2386 fits. Whereupon multiplying by 274 makes a unique solution.

[UncleEbenezer]

That last line of my post is of course wrong reasoning, right answer. The reasoning is 'cos I got the brackets wrong, then rationalised post-hoc from the answer that had emerged from the right reasoning!

[GoSeigen]

I came to the same answer with much the same reasoning as Uncle. Thanks again to Cinelli for posting these puzzles.

GS

[UncleEbenezer]

I came to the same answer with much the same reasoning as Uncle. Thanks again to Cinelli for posting these puzzles.

GS

Indeed, it's always a dilemma if you come to one of Cinelli's problems, and someone has already posted a solution. Do you go blind and post what's probably much the same, or defer to it? I've taken both approaches on different occasions.

[GoSeigen]

I came to the same answer with much the same reasoning as Uncle. Thanks again to Cinelli for posting these puzzles.

GS

Indeed, it's always a dilemma if you come to one of Cinelli's problems, and someone has already posted a solution. Do you go blind and post what's probably much the same, or defer to it? I've taken both approaches on different occasions.

Indeed. I had nothing interesting to add but didn't want Cinelli to get the impression no-one was following his posts...

GS

[cinelli]

Congratulations to the solvers and thank you for the kind words. Yes, I am always on the look out for new puzzles. I will write this reply in invisible ink in case anyone comes late to the party and wants to have a go at the puzzle.

Cinelli