Question

[Ngô Quốc Huy]

abcde=4.edcba

Answer 1

[phamthiminhtrang]

ABCDE 
x 4 
= EDCBA 

The answer is: 
21978 
x 4 
= 87912 

But rather than just give you the answer, here's how I figured it out. First, it is obvious that A must be an even number, because we are multiplying by 4 (an even number). The last digit will therefore be even. It can't be 0, because that would make ABCDE a four-digit number. It can't be more than 2, because that would result in a six-digit answer. So A = 2. 

2BCDE 
x 4 
= EDCB2 

So what can E be? The choices are E = {3, 8} because 3 x 4 = 12 and 8 x 4 = 32. But a value of 3 doesn't work in the result (3????) because it is too small. 

2BCD8 
x 4 
= 8DCB2 

Since the final number is 8 and we have 2 x 4, that means there is no carry from the prior multiplication (4 x B + carry). So B can't be anything higher than 1, possibly 0. 

Looking at the other side of the equation, we have 4D + 3 = (a number ending in 0 or 1). In other words, 4D must end in 7 or 8. Obviously only 8 works, because 4 is an even number. Working forward again, that means B = 1. 

21CD8 
x 4 
= 8DC12 

So what values of 4D result in a number ending 8? 4 x 2 = 8, 4 x 7 = 28. Now 2 is already taken and the problem said the digits were unique. So D = 7. 

21C78 
x 4 
= 87C12 

Finally, we have a carry of 3 (from 28 + 3 = 31). And when we calculate 4C + 3 it must also result in a carry of 3 and a last digit of C. In other words: 
4C + 3 = 30 + C 

This is easy to solve: 
3C = 27 
C = 9 

Thus the final answer is: 
21978 
x 4 
= 87912

Puzzling · 1 decade ago

Answer 2

[Super Xayda Vegito]

chả hiểu ji