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# Given that a, b, c, and d are different nonzero digits and

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Intern
Joined: 02 Sep 2004
Posts: 42

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Given that a, b, c, and d are different nonzero digits and [#permalink]

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17 Sep 2004, 07:33
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Given that a, b, c, and d are different nonzero digits and that 10d + 11c < 100 - a, which of the following could not be a solution to the addition problem below?

abdc
+ dbca

(A) 3689
(B) 6887
(C) 8581
(D) 9459
(E) 16091

Kudos [?]: 4 [0], given: 0

Director
Joined: 20 Jul 2004
Posts: 590

Kudos [?]: 158 [0], given: 0

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17 Sep 2004, 20:12
Good one.

10d + 11c < 100 - a means that the max value for a=9, b=9, c=7, d=8

IMO, Odd man out is (C) 8581 since
- to get a 5 in the hundredth digit, b+b has to be 7+7 and get a carry over from tenth digit
- to give a carry over of 1, d+c = 18, but the maximum possible value for d+c = 15

Hence C.

Kudos [?]: 158 [0], given: 0

Director
Joined: 05 May 2004
Posts: 574

Kudos [?]: 67 [0], given: 0

Location: San Jose, CA

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17 Sep 2004, 22:02
Got C

My approach is different:

a<100-10d-11c
or, a+c<100-10d-10c or a+c<10(10-d-c) ... A

Now a+c has to be >0
10-d-c>0 or d+c<10 .... hence we know for sure that adding d and c wont have a carry over

Also a+c can either be <10 or <20

if less than 10
a+c has no carry over and c+d has no carry over. Hence in the sum - b+b has to be even

if less than 20 and greater than 10
a+c has carryover of 1
10-d-c=2 ... from A
or d+c=8 .....
d+c+carryover_from_a+c=8+1=9
Hence b+b has no carryover to add ..... and hence b+b has to be even

Kudos [?]: 67 [0], given: 0

Director
Joined: 31 Aug 2004
Posts: 606

Kudos [?]: 147 [0], given: 0

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18 Sep 2004, 03:07
I think I have one more approach

abdc + dbca = 1000(a+d) +2.100b+ 10(d+c)+(c+a)

10d + 11c < 100 - a <=> 10(d+c)+(c+a) <100
and this represents tens+units digits of the sum
So there is no impact on the hundreds digit of the sum (=2.b)

(A) 3689
(B) 6887
(C) 8581
(D) 9459
(E) 16091

The sole solution with an odd number possibly equal to 2b is C with 5 or 15 which is impossible so answer si C

Kudos [?]: 147 [0], given: 0

Joined: 31 Dec 1969

Kudos [?]: [0], given:

Location: Russian Federation
Concentration: Entrepreneurship, International Business
GMAT 3: 740 Q40 V50
GMAT 4: 700 Q48 V38
GMAT 5: 710 Q45 V41
GMAT 6: 680 Q47 V36
GMAT 9: 740 Q49 V42
GMAT 11: 500 Q47 V33
GMAT 14: 760 Q49 V44
WE: Supply Chain Management (Energy and Utilities)

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19 Sep 2004, 23:50
twixt wrote:
I think I have one more approach

abdc + dbca = 1000(a+d) +2.100b+ 10(d+c)+(c+a)

10d + 11c < 100 - a <=> 10(d+c)+(c+a) <100
and this represents tens+units digits of the sum
So there is no impact on the hundreds digit of the sum (=2.b)

(A) 3689
(B) 6887
(C) 8581
(D) 9459
(E) 16091

The sole solution with an odd number possibly equal to 2b is C with 5 or 15 which is impossible so answer si C

I follow your point up to where you have the bold statement. could you elaborate how you selected C as the odd one out from there, Why 5 or 15

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Director
Joined: 31 Aug 2004
Posts: 606

Kudos [?]: 147 [0], given: 0

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20 Sep 2004, 03:22
Guest,

The hundreds number of the sum could have several origins :

we know that it is at least equal to the units digit of 2.b (2.b could be above or below 10...)
you could have to add +1 if 10(d+c)+(c+a) >99 but we know from the stem that it is < 100 so only 2.b is contributing to the hundreds digit of the sum ; 2.b beeing even, the sole not compliant number is C

Kudos [?]: 147 [0], given: 0

20 Sep 2004, 03:22
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# Given that a, b, c, and d are different nonzero digits and

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