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More cryptarithms - Lucky 7's fixed


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12 replies to this topic

#1 bonanova

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Posted 10 September 2012 - 10:35 PM

The prime multiplication was solved quickly.
Congratulations to curr3nt and Rob_Gandy, the early solvers.
These may pose a better challenge; a three-pack!

Letter clues constrain the substituted numbers: same digit for every A, etc.
Asterisks are wild cards, can be any digit.

1. Easy as A-B-C.

. . . A B C
. . . B A C
. . -------
. . * * * *
. . * * A
* * * B
-----------
* * * * * *



2. Four Aces.

. . . * * *
. . . * 2 *
. . . -----
. . . * * *
. * * * *
. * 8 *
-----------
* * 9 * 2 *



3. Lucky 7's.
. . . . . . . . . . . * * 7 * *
. . . . . .--------------------
* * * 7 * / * * 7 * * * * * * *
. . . . . . * * * * * *
. . . . . . -------------
. . . . . . * * * * 7 7 *
. . . . . . * * * * * * *
. . . . . . -------------
. . . . . . . . * 7 * * * *
. . . . . . . . * 7 * * * *
. . . . . . . . -------------
. . . . . . . . * * * * * * *
. . . . . . . . * * * * 7 * *
. . . . . . . . ---------------
. . . . . . . . . . * * * * * *
. . . . . . . . . . * * * * * *


Edit: the leftmost number needs a leading asterisk.
It should read: * * * * 7 *
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#2 Rob_Gandy

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Posted 10 September 2012 - 11:01 PM

Spoiler for ABC?

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#3 bonanova

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Posted 10 September 2012 - 11:52 PM

Spoiler for ABC?


Got it. Nice.
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#4 Rob_Gandy

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Posted 11 September 2012 - 12:07 AM

Spoiler for Four Aces

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#5 curr3nt

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Posted 11 September 2012 - 04:08 PM

The Lucky 7 is impossible. How can a five digit number times a single digit yield a seven digit number?

. . . . . . . . . . . * * 7 * *
. . . . . .--------------------
* * * 7 * / * * 7 * * * * * * *
. . . . . . * * * * * *
. . . . . . -------------
. . . . . . * * * * 7 7 *
. . . . . . * * * * * * *
. . . . . . -------------
. . . . . . . . * 7 * * * *
. . . . . . . . * 7 * * * *
. . . . . . . . -------------
. . . . . . . . * * * * * * *
. . . . . . . . * * * * 7 * *
. . . . . . . . ---------------
. . . . . . . . . . * * * * * *
. . . . . . . . . . * * * * * *

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#6 bonanova

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Posted 11 September 2012 - 05:00 PM

Sorry. It's division this time.
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- Bertrand Russell

#7 curr3nt

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Posted 11 September 2012 - 06:17 PM

Fifth row of numbers should be ***7* times the second number in **7**. The result is a seven digit number which isn't possible.

I would expect an equation like below otherwise you are doing division much differently than I was taught. The remainder of each step should be less than divisor. If the remainder is six digits like you have listed in two steps then I have no idea since a six digit number is bigger than a five digit divisor.

. . . . . . . . . . . * * 7 * *
. . . . . .--------------------
* * * 7 * / * * 7 * * * * * * *
. . . . . . * * * * * *
. . . . . . -------------
. . . . . .. * * * 7 7 *
. . . . . . . * * * * * *
. . . . . . -------------
. . . . . . . . * 7 * * * *
. . . . . . . . * 7 * * * *
. . . . . . . . -------------
. . . . . . . . . * * * * * *
. . . . . . . . . * * * 7 * *
. . . . . . . . ---------------
. . . . . . . . . . * * * * * *
. . . . . . . . . . * * * * * *

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#8 CaptainEd

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Posted 11 September 2012 - 06:18 PM

(edit--Curr3nt has now answered better than I was going to)

Edited by CaptainEd, 11 September 2012 - 06:19 PM.

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#9 curr3nt

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Posted 11 September 2012 - 06:21 PM

Does this better illustrate the issue I'm having?

. . . . . . . . . . . A B 7 C D
. . . . . .--------------------
W X Y 7 Z / * * 7 * * * * * * *
. . . . . . * * * * * * . . . . .WXY7Z * A
. . . . . . -------------
. . . . . . * * * * 7 7 *
. . . . . . * * * * * * * . . . .WXY7Z * B (How can this be 7 digits?)
. . . . . . -------------
. . . . . . . . * 7 * * * *
. . . . . . . . * 7 * * * * . . .WXY7Z * 7
. . . . . . . . -------------
. . . . . . . . * * * * * * *
. . . . . . . . * * * * 7 * * . .WXY7Z * C (How can this be 7 digits?)
. . . . . . . . ---------------
. . . . . . . . . . * * * * * *
. . . . . . . . . . * * * * * * .WXY7Z * D

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#10 bonanova

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Posted 13 September 2012 - 07:59 AM

You're exactly right. The leftmost number needs another leading digit.

V W X Y 7 Z

Then it should be solvable. I've edited the OP as well.
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- Bertrand Russell




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