AB CD+ AAA= where AB and CD are two-digit numbers and

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AB CD+ AAA= where AB and CD are two-digit numbers and [#permalink]

06 May 2007, 03:37

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AB
CD+
AAA=

where AB and CD are two-digit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C?

(A) 1

(B) 3

(C) 7

(D) 9

(E) Cannot be determined

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[#permalink]

10 Sep 2007, 14:40

Get 9 too

we know that A is 1...

so

1B
CD
111

try a couple of number 15 for AB and 96 for CD

you can vary B and D but C is always 9..

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Re: PS = Addition [#permalink]

10 Sep 2007, 15:57

Mishari wrote:

I think 9
A=1 since anything less than 100 add together will be less than 200

So if B+D <= 9, then
B+D = 1
1+C = 11
C=10

if B+D >9, then
B+D = 11
1+1+C=11
C=9

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Re: PS = Addition [#permalink]

10 Sep 2007, 16:50

Mishari wrote:

everyone else seemed to get 9. C too

18 + 93 = 111

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[#permalink]

10 Sep 2007, 18:48

Kalyan wrote:

bkk145/ fresinha12, why is A=1 only?

Because if you add two digit numbers, the result must be less than 200
you also know that the result has 3 digits. Therefore, it must be that
100 <= AAA <=199
Therefore, A can only be 1.

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Re: PS = Addition [#permalink]

10 Sep 2007, 19:32

Mishari wrote:

Aha finally got it.

Since AB and CD are two digit numbers and AAA is a three digit number. AAA has to be less than 200. Ex/ 99+99=198. So since the last digit of AAA has to be 1 then all the A's are 1.

So the new problem looks like this:

1B
+CD
111

To get 1 from B+D, B+D has to either equal 11 or 1 (0+1). In this case only 11 works b/c C can't be 10.

It doesnt matter what the values of B and D are, but just that they equal 11. Since there is a carry over of 1. C must be equal to 9. Since 9+1+1=11.

The numbers could look like this.

15
+96
111

Ans D

Re: PS = Addition [#permalink] 10 Sep 2007, 19:32

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AB CD+ AAA= where AB and CD are two-digit numbers and

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AB CD+ AAA= where AB and CD are two-digit numbers and

AB CD+ AAA= where AB and CD are two-digit numbers and

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