Another Silly Riddle for you guys

[DaShen]

ABCD * 4 = DCBA

A = ?
B = ?
C = ?
D = ?

It is easy once you think about it. But it is fun to think about this kind of stuff. And for you CS guys (no it is not Hexadecimal).

BTW... it is not 0000 * 4 = 0000
It is actual numbers other than zero.

 

[her209]

ABCD = 0000

 

[chrisms]

0

 

[DaShen]

Originally posted by: her209
ABCD = 0000

Although it works, there is a more elegant solution.

 

[eLiu]

2178

edit: 4*D = A mod 10. Also A<3 else you get 5 digits. So A = 2, D = 8 is the only choice.

Then (4*C+3) = B mod 10 because 8*4=32, so you carried a 3. Again B < 2 else the carry-over will change the 1000s place.

From there, C = 7, B = 1.

 

[Leros]

As eLiu pointed out, the other answer is 2178.

I just wrote a program to test all possible solutions. Seemed easier than thinking about it.

 

[her209]

2178,8712

Doh! Too late.

 

[thesurge]

Extension:
AB....4*4=4AB.... (Ex. AB.....4--->4AB.... as 1234--->4123)

AB..... and 4AB....each represent a distinct integer (with digits A,B etc.).

find AB....

 

[jordanz]

1111

 

[Leros]

Originally posted by: jordanz
1111

fail

4444 != 1111

 

[dlock13]

This is the kind of question we get on these lame "Math League" tests. There are six questions which seem hard, but once given the solution, they are rather simple.

I personally like the challenge.

 

[DaShen]

Originally posted by: eLiu
2178

edit: 4*D = A mod 10. Also A<3 else you get 5 digits. So A = 2, D = 8 is the only choice.

Then (4*C+3) = B mod 10 because 8*4=32, so you carried a 3. Again B < 2 else the carry-over will change the 1000s place.

From there, C = 7, B = 1.

Correct sir.

BUMP for work crowd

 

[George P Burdell]

Nice