[bonanova]

Here's a long division problem.

There is a 7 in the quotient.

Can you find the other numbers?

.....___X7XXX

XXX /XXXXXXXX

.....XXXX

.......XXX

.......XXX

.......XXXX

....... XXX

.........XXXX

.........XXXX

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

[Guest]

.....___X7X0X

1XX /XXXXXXXX

.....XXXX

.......XXX

.......XXX

.......1XXX

....... XXX

.........XXXX

.........XXXX

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

Edited January 9, 2009 by tbrophy

[Guest]

.....___X7X0X

1XX /XXXXXXXX

.....XXXX

.......XXX

.......XXX

.......1XXX

....... XXX

.........XXXX

.........XXXX

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

.....___(8or9)7X0(8or9)

1XX /1XXXXXXX

.....1XXX

.......(8or9)XXX

.......(7or8)XX

.......1XXX

....... XXX

.........1XXX

.........1XXX

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

[bonanova]

Yes! You're both on the right track exactly.

[Guest]

Thanks I should have gotten the 1s on the 4 digit numbers

. On the 8's or 9's on the top number, wouldn't anything 5 or higher be acceptable?

.....___(8or9)7X0(8or9)

1XX /1XXXXXXX

.....1XXX

.......(8or9)XXX

.......(7or8)XX

.......1XXX

....... XXX

.........1XXX

.........1XXX

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

[Guest]

Thanks I should have gotten the 1s on the 4 digit numbers

. On the 8's or 9's on the top number, wouldn't anything 5 or higher be acceptable?

you know that a 7 yields a product of three digits, so to get a product of 4 digits you need an 8 or 9

Edited January 9, 2009 by Cherry Lane

[Guest]

Well... I can fill in the blanks, but I don't have unique digits in the answer..

12040470/123=97890 you can fill in the blanks then.

[bonanova]

Well... I can fill in the blanks, but I don't have unique digits in the answer..

12040470/123=97890 you can fill in the blanks then.

Not quite ...

[Guest]

.....___L7MNO

ABC /DEFGHIJK DEFG = 1099-1251, ABC=112-128

.....XXXX LxABC value is 1000-1152

.......XXX DEFG-XXXX value is 884-999

.......XXX 7xABC value is 784-899

.......XXXX first digit = 1, >1000

....... XXX M*ABC=3 digit, >899

.........XXXX

.........XXXX

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

1) 7 x ABC = 3 digits with 3 digit remainder so 101 < ABC < 128

2) ABC *L is 4 digits so L = 8 or 9 and 112 <ABC<128

3) using ABC * 8 or 9 get range of 1000-1152 [1116] for first XXXX

4) DEFG – XXXX <=99 so DEFG = 1099-1251 [1215]

5) M*ABC = 3 digit, >899 (to leave 2 digit remainder) so M = 8

6) 8*ABC 3 digit limits 112<ABC<124 [RECALC 2]

7)

8)

.....___L78NO

1BC /1EFGHIJK DEFG = 1099-1215, ABC=112-124

.....1XXX LxABC value is value is 1000-1116

.......XXH DEFG-XXXX value is 884-968

.......XXX 7xABC value is value is 784-868

.......1XXI first digit = 1, >1000

....... XXX M*ABC=3 digit, >899

.........XXJK

.........XXJK NO*ABC = 4 digits

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

What I've gotten so far... fun but very time consuming puzzle

<_<

[Guest]

I haven’t given up on this! I looked at it again this morning, as a wake-me-up, and here’s where I ended up:

.....___X7XXX

XXX /XXXXXXXX

.....XXXX

.......XXX

.......XXX

.......XXXX

........XXX

.........XXXX we know from this line the fourth digit of quotient =0

.........XXXX

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

.....___A7H0M

BXX /CXXXXXXX ...... 7 x BXX yields a three-digit number (FXX), so

.....DXXX .......... B = 1, F = 7,8,or 9

.......EXX ......... A x BXX = DXXX (4 digits), so A > 7, D=1 and C=1

.......FXX ......... also, MxBXX = LXXX (4 digits), so M > 7 L=1

.......GXXX ........ (also K=1)

........JXX ........ EXX – FXX = GXX

.........KXXX ...... E>F, F=7 or 8, E=8 or 9

.........LXXX ...... GXXX-JXX=KX, so G=1

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

.....___A7H0M

1XX /1XXXXXXX ..... 1XXX-JXX=1X (2 digits), so J=9

.....1XXX ......... Hx1XX=JXX=9XX > FXX (above, 7<=F<=8)

.......EXX ........ so H=8, A=9, M=9

.......FXX

.......1XXX

........JXX

.........1XXX

.........1XXX

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

.....___97809

1XX /1XXXXXXX ..... 8x1XX = 9XX

.....1XXX ......... 112<1XX<125

.......EXX

.......FXX

.......1XXX

........9XX

.........1XXX

.........1XXX

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

After this, I don’t know what to do except trial-and-error for the divisor. 9x1XX must yield a remainder (EX) large enough that EXX-FXX>=100. Using the range 112<1XX<125, the only one that works is 124. So:

.....___97809

124 /12128316

.....1116

.......986

.......868

.......1003

........992

.........1116

.........1116

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

But I’d love for some help in getting this last bit logically!

[bonanova]

I haven’t given up on this! I looked at it again this morning, as a wake-me-up, and here’s where I ended up:

.....___X7XXX

XXX /XXXXXXXX

.....XXXX

.......XXX

.......XXX

.......XXXX

........XXX

.........XXXX we know from this line the fourth digit of quotient =0

.........XXXX

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

.....___A7H0M

BXX /CXXXXXXX ...... 7 x BXX yields a three-digit number (FXX), so

.....DXXX .......... B = 1, F = 7,8,or 9

.......EXX ......... A x BXX = DXXX (4 digits), so A > 7, D=1 and C=1

.......FXX ......... also, MxBXX = LXXX (4 digits), so M > 7 L=1

.......GXXX ........ (also K=1)

........JXX ........ EXX – FXX = GXX

.........KXXX ...... E>F, F=7 or 8, E=8 or 9

.........LXXX ...... GXXX-JXX=KX, so G=1

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

.....___A7H0M

1XX /1XXXXXXX ..... 1XXX-JXX=1X (2 digits), so J=9

.....1XXX ......... Hx1XX=JXX=9XX > FXX (above, 7<=F<=8)

.......EXX ........ so H=8, A=9, M=9

.......FXX

.......1XXX

........JXX

.........1XXX

.........1XXX

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

.....___97809

1XX /1XXXXXXX ..... 8x1XX = 9XX

.....1XXX ......... 112<1XX<125

.......EXX

.......FXX

.......1XXX

........9XX

.........1XXX

.........1XXX

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

After this, I don’t know what to do except trial-and-error for the divisor. 9x1XX must yield a remainder (EX) large enough that EXX-FXX>=100. Using the range 112<1XX<125, the only one that works is 124. So:

.....___97809

124 /12128316

.....1116

.......986

.......868

.......1003

........992

.........1116

.........1116

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

But I’d love for some help in getting this last bit logically!

Gold star [and red badge of courage] for CL.

This problem might be like the extra-tough SUDOKU problems that do require some trial and error.

You bracketed the answer to a searchable size, and that might be the best you can do.

[Guest]

This was great!

I got GXXX to 10XX before I started trying out the range of numbers.

Very nice puzzle indeed.