+0  
 
0
1
283
1
avatar

If ABCDE x 4 = EDCBA what is the number ABCDE? As usual with this type of problem each letter is a digit and different letters represent different digits.

 Jun 24, 2018
 #1
avatar+2340 
+1

This problem just requires some logic to crack it. 

 

Both numbers have 5 digits. A must be equal to or equal to or less than 2 because, otherwise, the result would exceed the 5-digit restriction. 4E is even, so A only has one value: 2. This leaves us with 2BCDE*4=EDCB2

 

4E must end in a 2. Only two values achieve this: 3 and 8. E cannot be 3 because 20000*4=80000. 8 is the only option then. This leaves us with 2BCD8*4=8DCB2

 

B2 must be a multiple of 4 because of the divisibility rules for 4. B cannot exceed 2 because 3000*4=12000. This would cause a carry for 2BCD8*4, and the result would be impossible. Only 1 works because that makes 8DCB2 end in a 12, which is divisible by 4. This leaves us with 21CD8*4=8DC12

 

4D+3 must end in a 1, so 7 and 1 are the only options. D cannot be 1 because all the digits must be unique, according to the original problem; therefore, d=7. This leaves us with 21C78*4=87C21

 

We have a carry of 3 from 4*7+3=31, and 4c+3 must result in a carry of 3 so that the result holds true. This leaves us with a basic algebra problem:

 

\(4c+3=c+30\) 
\(3c=27\) 
\(c=9\) 
  

 

c=9

 

Therefore, \(21978*4=87912\)

.
 Jun 24, 2018
edited by TheXSquaredFactor  Jun 24, 2018

11 Online Users

avatar
avatar
avatar