More cryptarithms - Lucky 7's fixed

[bonanova]

The 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 *

[Rob_Gandy]

1. Easy as A-B-C.

. . . 2 8 6

. . . 8 2 6

. . -------

. . 1 7 1 6

. . 5 7 2

2 2 8 8

-----------

2 3 6 2 3 6

[bonanova]

1. Easy as A-B-C.

. . . 2 8 6

. . . 8 2 6

. . -------

. . 1 7 1 6

. . 5 7 2

2 2 8 8

-----------

2 3 6 2 3 6

Got it. Nice.

[Rob_Gandy]

2. Four Aces. (11*11=121 'four aces' very nice)

. . . 9 8 7

. . . 1 2 1

. . . -----

. . . 9 8 7

. 1 9 7 4

. 9 8 7

-----------

1 1 9 4 2 7

At first I thought the title meant there were 4 1's in the factors.

[curr3nt]

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 * *

. . . . . . . . ---------------

. . . . . . . . . . * * * * * *

. . . . . . . . . . * * * * * *

[bonanova]

Sorry. It's division this time.

[curr3nt]

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 * *

. . . . . . . . ---------------

. . . . . . . . . . * * * * * *

. . . . . . . . . . * * * * * *

[CaptainEd]

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

Edited September 11, 2012 by CaptainEd

[curr3nt]

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

[bonanova]

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.

[curr3nt]

Figuring the 5878* and 12547* was fairly easy. I did trial and error to get the last two *.

. . . . . . . . . . . . 5 8 7 8 1

. . . . . . .--------------------

1 2 5 4 7 3 / 7 3 7 5 4 2 8 4 1 3

. . . . . . . 6 2 7 3 6 5

. . . . . . . -------------

. . . . . . . 1 1 0 1 7 7 8

. . . . . . . 1 0 0 3 7 8 4

. . . . . . . -------------

. . . . . . . . . 9 7 9 9 4 4

. . . . . . . . . 8 7 8 3 1 1

. . . . . . . . . -------------

. . . . . . . . . 1 0 1 6 3 3 1

. . . . . . . . . 1 0 0 3 7 8 4

. . . . . . . . . ---------------

. . . . . . . . . . . 1 2 5 4 7 3

. . . . . . . . . . . 1 2 5 4 7 3

[Rob_Gandy]

Figuring the 5878* and 12547* was fairly easy. I did trial and error to get the last two *.

. . . . . . . . . . . . 5 8 7 8 1

. . . . . . .--------------------

1 2 5 4 7 3 / 7 3 7 5 4 2 8 4 1 3

. . . . . . . 6 2 7 3 6 5

. . . . . . . -------------

. . . . . . . 1 1 0 1 7 7 8

. . . . . . . 1 0 0 3 7 8 4

. . . . . . . -------------

. . . . . . . . . 9 7 9 9 4 4

. . . . . . . . . 8 7 8 3 1 1

. . . . . . . . . -------------

. . . . . . . . . 1 0 1 6 3 3 1

. . . . . . . . . 1 0 0 3 7 8 4

. . . . . . . . . ---------------

. . . . . . . . . . . 1 2 5 4 7 3

. . . . . . . . . . . 1 2 5 4 7 3

How......? I had no clue where to even start on this one. What was your process for finding the numbers?

[curr3nt]

Let's see...using VWXY7Z and AB7CE and numbering the steps 1 through 5. (A * VWXY7Z = step 1)

1. V has to be 1 and B and C have to be 8 or 9. That is the only way VWXY7Z * 7 = 6 digits and VWXY7Z * B (or C) = 7 digits.

2. Step 3 has to start with 9 and 8 or 8 and 7 which lead me to find VWXY7Z = 124Y7Z or 125Y7Z for small sets of Y. I ended up finding 12547Z fit using step 4's second number as a template.

3. C = 8 was then the only only that would work for step 4

4. Looked at B = 8 or 9 for step 2 can't remember what eliminated B = 9

5. Once I knew what the third digit of the first number in step 2 was I figured out what the third digit of the second number in step 1 would need to be. Only A = 3 or 5 would give me the number I wanted.

6. Looking at 125470 * 38780 and 125479 * 38789 eliminated A= 3

Then it was trial and error for Z and D