Reverse 5 digit number by multiplying 4

[Rakesh Joshi]

A, B, C, D and E are 5 different single digit numbers. The number ABCDE times 4 will have a result of EDCBA. What digit numbers do A, B, C, D, and E represent?

ABCDE
* 4
-------
EDCBA
-------

[Garrett Rowe]

21978

[Shaan Shar]

Could any one tell me what is the logic to calculate this problem. Atleast direct others how to calculate this problem or tell the algorithm to solve it....

[Garrett Rowe]

Originally posted by Ankur Sharma:
Could any one tell me what is the logic to calculate this problem. Atleast direct others how to calculate this problem or tell the algorithm to solve it....

21978 * 4 = 87912

I just used a simple loop from 10000 to 99999 and multiplied each number by 4, converted to a String, and checked to see if the String reversed was equal to the original number as a String.

[Ryan McGuire]

Originally posted by Ankur Sharma:
Could any one tell me what is the logic to calculate this problem. Atleast direct others how to calculate this problem or tell the algorithm to solve it....

You could look at it as an integer linear programming problem.

Find A,B,C,D,E such that.

A*10000 + B*1000 + C*100 + D*10 + E = 4*(E*10000 + D*1000 + C*100 + B*10 + A)
- or -
A*9996 + B*960 - C*300 - D*3990 - E*39999 = 0
A >= 0
A <= 9
B >= 0
B <= 9
C >= 0
C <= 9
D >= 0
D <= 9
E >= 0
E <= 9

[fred Joly]

- A can only be 1 or 2
if A > 2 then the answer would be of the form FEDCBA


-let's try with A = 1
1BCDE
*4
------
EDCB1

impossible : 4*E = some number ending by 1

- then A "MUST" be 2
2BCDE
*4
-----
EDCB2


- E=3 or E=8 (E*4 has to be a number ending with 2)

let's try E=8 (looks like the easier...)

2BCD8
*4
-----
8DCB2


- B can only 1 or 2 (otherwise 4*B > 10 and that's now impossible)
Since we already have a 2 , b=1

21CD8
*4
-----
8DC12


- we can write : D*4 + 3 = some number ending by 1
->D*4 = some number ending by 8

D=2 or D=7 => D=7 (A = already 2)


21C78
*4
-----
87C12


- (4*21000 = 84000)
we can write

C78
*4
-----
3C12

we can write C*4 + 3 = 3C (beware 3C means thirty C)
so C >= 7
we already have a 7 and a 8 => C = 9

21978
*4
-----
87912


No computer needed to do this...

[Vishnu Prakash]

   
 
public class Check {
public static void main(String[] args) {
for(int i=10000; i<99999; i++) {
 
int a = i;
int b = i*4;
 
Integer x = new Integer(a);
Integer y = new Integer(b);

String in1 = x.toString();
String in2 = y.toString();
 
//System.out.println("Values: " + in1 + " : " + in2);

StringBuffer m = new StringBuffer(in1);
StringBuffer n = new StringBuffer(in2);
 
StringBuffer o =  n.reverse();
 
String st1 = new String(m);
String st2 = new String(o);
 
Integer ans1 = new Integer(st1);
Integer ans2 = new Integer(st2);
 
int final1 = ans1.intValue();
int final2 = ans2.intValue();

if(final1==final2){
System.out.println("\n\nAnswer:" + final1 + " : " + (final1*4) + "\n\n");
break;
}
}
}
}
 

[Stefan Wagner]

A great performance-improvement would be to reduce the loop to unique digits in the numbers:

for(int i=12345; i < 98766; ++i)

[Edward Collins]

Microsoft's VBA:

For x = 10000 To 25000
   a$ = Trim(Str(x))
   Reverse_a$ = ""
   For y = Len(x) To 1 Step -1
       Reverse_a$ = Trim(Reverse_a$ + Mid(a$, y, 1))
   Next
   If Reverse_a$ = Trim(Str(x * 4)) Then MsgBox (a$)
Next x

[Campbell Ritchie]

Welcome to the Ranch That looks good for a solution with a loop. I trust it took you less than 15 years to find