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