# hi new puzzle for u guys

• May 3rd 2007, 01:29 PM
ggw
hi new puzzle for u guys
Starting with the two two-digit numbers
ab and cd, where ab > cd,
concatenate them to obtain the
four-digit number abcd.
(note: each of the numbers: a,b,c,d are digits)

If you subtract the difference between
ab and cd from abcd, you obtain 5689.

Determine the number cd.

Last puzzle that I posted. There is no one solve it. I will post solution up this weekend...any suggestions ? and here is a little bit easier puzzle
• May 3rd 2007, 03:27 PM
ecMathGeek
Quote:

Originally Posted by ggw
Starting with the two two-digit numbers
ab and cd, where ab > cd,
concatenate them to obtain the
four-digit number abcd.
(note: each of the numbers: a,b,c,d are digits)

If you subtract the difference between
ab and cd from abcd, you obtain 5689.

Determine the number cd.

Last puzzle that I posted. There is no one solve it. I will post solution up this weekend...any suggestions ? and here is a little bit easier puzzle

We have a number: a*10^3 + b*10^2 + c*10 + d

If we add: c*10 + d, we get:
a*10^3 + b*10^2 + 2c*10 + 2d

If we subract: a*10 + b, we get:
a*10^3 + b*10^2 + (2c - a)*10 + (2d - b) = 5689

b >= 1 --> a = 5
b - 1 = 6 --> b = 7
2d + 10 - b = 9 --> d = 3
2c + 10 - a - 1 = 8 --> c = 2

10c + d = 23
• Jun 1st 2007, 04:18 AM
ahmed1313
ab > cd
abcd-(ab-cd)=5689
cd=?

solution:

let a=5, b=7 or 6
let cd=X

(1) for a=5, b=7

(5700+X)-(57-X)=5689
X=cd=23

(2) for a=5, b=6
(5600+X)-(56-X)=5689