ABCDE * 4 = inference process of edcba

It is known that numbers A, B, C, D, and E are 0-9 and meet the requirements of ABCDE * 4 = EDCBA. Please solve the problem.

 

Analysis process:

 

1. because both are 5 digits, A can only be 1 or 2, and because 4 on the left multiplied by any number is an even number, so the right one (A) must be an even number, we can see that A = 2, in this case, the original format is: 2 BCDE * 4 = EDCB2

 

2. A = 2, so the tens of thousands of E> = 8 (Conclusion 2.1) on the Right; E * 4 = B2 on the left, so E can only be 3 or 8 (Conclusion 2.2 ), combined with conclusion 2.1 and Conclusion 2.2, we can see that E = 8. In this case, the original formula is: 2BCD8*4 = 8DCB2.

 

3. E = 8. Therefore, the left-side kilobytes of B * 4 cannot be carried, that is, B can only be 0, 1, 2 (Conclusion 3.1), and the left-side digits of 8*4 = 32

If B = 0, the number of digits in the left ten D * 4 must be 7, which is impossible;

If B = 2, the number of digits in the left ten D * 4 must be 9, which is not possible;

So B = 1 (this can also be explained, because the single digit of D * 4 [that is, the ten digits on the right] must be an even number, and 8*4 = 32, 3 must be added with an odd number to be equal to the even number, but among the numbers 0, 1 and 2, only 1 is an odd number, so B = 1)

In this case, the original format is: 21CD8*4 = 8DC12

 

4. the first digit on the left is 8*4 = 32, and the second digit (B) On the right is 1. Therefore, the first digit of D * 4 should be 11-3 = 8, so D can only be 2 or 7, replace D = 2 and D = 7 into the original formula, and then we can infer: D = 7, C = 9

 

So far, the results of A, B, C, D, and E have been inferred, namely, 2, 1, 9, 7, and 8, respectively.

 

Result:

21978*4 = 87912