PS : Digits : PS Archive
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# PS : Digits

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CEO
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13 Sep 2003, 05:00
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Do this in 1.5 mins and explain at leisure

In a 4-digit number, the sum of the first two digits is equal to that
of the last two digits. The sum of the first and last digits is equal to the third digit. Finally, the sum of the second and fourth digits is twice the sum of the other two digits. What is the third digit of the number?

a) 5
b) 8
c) 1
d) 4
Senior Manager
Joined: 22 May 2003
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13 Sep 2003, 20:33
Anser is A. Took me like 4 min

If number is : ABCD

1)A+B=C+D
2)A+D=C
3)B+D=2A+2C

From 1 and 2: 2A+B=2C
Combine this with (3) and you get D=4A

So (using 2) C=5A

Since 1<=A<=9 and C=5A C could only be 1
SVP
Joined: 30 Oct 2003
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18 Jan 2004, 22:22
Answer is 4. It took me 2 mins
a,b,c,d are the digits
1. a + b = c + d
2. a + d = b
3. b + d = 2(a+c)

from 1 and 2 we have
2a = c

from 1 b = d+c/2 or b-d = c/2
from 3 we have 2b = 3c+c/2 so b = 7c/4

From this we know that c = 4 and b = 7
Director
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18 Jan 2004, 23:52
Though not a choice, zero also works.

I think setting up the equations was the easy part.

I wonder if there's a quick, methodical and easy way to solve the equations.
SVP
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19 Jan 2004, 05:20
Yeah u got me there. The problem stem should be changed to distinct different digits.
Intern
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20 Jan 2004, 14:40
Since a is the first digit of a 4 digit number, it is not zero.

Given the explanations above, the correct answer is A) C= 5
a<>0   [#permalink] 20 Jan 2004, 14:40
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