a, b, c, d and e are all non-zero distinct single digit integers. What

Question

a b c d e
 x 4
-----------
e d c b a

a, b, c, d and e are all non-zero distinct single digit integers. What are the values of a, b, c, d and e?

cipher · Answer

Okay, let me give it a shot to this one...! I am assuming that a,b,c,d,e are distinct integers..!

First thing to note the 5 digit number is multiplied by 4 and gives a 5 digit number again .. that means abcde has to be equal or less than 24999 else the abcde when multiplied by 4 will become 6 digit number.

Given:
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

gauravnagpal · Answer

I am sure it can be DS question if not the PS one !!1....tricky time consuming 
is there any other way ...?

ykaiim · Answer

No, I think nitish explained well.

KarishmaB · Answer

A 5 digit number multiplied by 4 gives another 5 digit number. So a will be either 1 or 2. 
edbca has 4 as a factor so a will be 2.
4 times 2 b c d e will begin with 8 or 9 so e = 8 or 9. But 8 times 4 will give a units digit of 2. So e = 8.

2 b c d 8
........* 4
8 d c b 2

So now you have
b c d(+3)
....*4
d c b

4 times a three digit number gives another 3 digit number. So b must be 1. 
So 4d + 3 =_ 1
4d = _ 8
d = 7

2 1 c 7 8
..........*4
8 7 c 1 2

4c + 3 = 3 c
c = 9

abcde = 21978

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a, b, c, d and e are all non-zero distinct single digit integers. What

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