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

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Director
Joined: 24 Aug 2007
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a, b, c, d and e are all non-zero distinct single digit integers. What  [#permalink]

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Updated on: 09 May 2010, 08:10
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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?

Originally posted by ykaiim on 09 May 2010, 05:40.
Last edited by ykaiim on 09 May 2010, 08:10, edited 1 time in total.
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Joined: 25 Jun 2009
Posts: 211
Re: a, b, c, d and e are all non-zero distinct single digit integers. What  [#permalink]

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09 May 2010, 08:01
1
2
ykaiim wrote:
a b c d e
x 4
-----------
e d c b a
-----------

What are the values of a, b, c, d and e?

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

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

21978
x 4
= 87912
Manager
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Re: a, b, c, d and e are all non-zero distinct single digit integers. What  [#permalink]

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27 Aug 2010, 05:43
1
I am sure it can be DS question if not the PS one !!1....tricky time consuming
is there any other way ...?
Director
Joined: 24 Aug 2007
Posts: 598
WE 1: 3.5 yrs IT
WE 2: 2.5 yrs Retail chain
Re: a, b, c, d and e are all non-zero distinct single digit integers. What  [#permalink]

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27 Aug 2010, 19:53
No, I think nitish explained well.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10474
Location: Pune, India
Re: a, b, c, d and e are all non-zero distinct single digit integers. What  [#permalink]

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27 Jun 2016, 20:51
ykaiim wrote:
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 intergers.

What are the values of a, b, c, d and e?

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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Re: a, b, c, d and e are all non-zero distinct single digit integers. What  [#permalink]

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15 May 2020, 07:01
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Re: a, b, c, d and e are all non-zero distinct single digit integers. What   [#permalink] 15 May 2020, 07:01