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AB and BA are 2-digits integers and AB*BA=CDE, where CDE is

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AB and BA are 2-digits integers and AB*BA=CDE, where CDE is [#permalink] New post 01 Apr 2008, 16:43
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A
B
C
D
E

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AB and BA are 2-digits integers and AB*BA=CDE, where CDE is a 3-digits integer. What is AB?
1). A=E
2). A=C
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Re: DS problem solving [#permalink] New post 01 Apr 2008, 18:20
I get E..

AB can be 11

then you get CDE=121 meets both

AB=21
CDE=252 meets both statements

Insuff
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Re: DS problem solving [#permalink] New post 01 Apr 2008, 18:32
I get E as well.

A=E could be 21*12=152 and 13*31=403 so insuff

A=C Could be 11*11 or 12*12 so insuff
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Re: DS problem solving [#permalink] New post 01 Apr 2008, 19:08
young_gun wrote:
AB and BA are 2-digits integers and AB*BA=CDE, where CDE is a 3-digits integer. What is AB?
1). A=E
2). A=C


I will go for E. below is my reasoning

AB
BA
--
CDE

(1)
AB
BA
----
CDA => A*B = A => B = 1 or A= 0 => since BA is 2 digit integer A can not be 0 => so B =1

(2)
AB
BA
----
ADE => not suff

combining

AB
BA
-----
ADA => (A^2+ b^2 = D)still does not tell anything about A
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Re: DS problem solving [#permalink] New post 01 Apr 2008, 22:03
young_gun wrote:
AB and BA are 2-digits integers and AB*BA=CDE, where CDE is a 3-digits integer. What is AB?
1). A=E
2). A=C


we know AB and BA are two digit integers so 0 cannot be any of the digit.
also we know CDE is a 3 digit integer so it is < than 1000 but > than 121. so:-

i. digits can only be 1, 2, 3, 4, 5, and 6.
ii. 1000 > CDE > 121.

from 1: AB can be of any value from 11, 12, 13, 14, 15, 16, 21, 31 41, 51, and 61. any value of AB results in A=E if unit digit of AB is 1. so nsf.
from 2: again AB can be of any value from 11, 12, 13, 14, 15, 16, 21, 31 41, 51, and 61. but to have A=C, AB must be 21.

if AB is 13, ABxBA = 403. so C=4, which is not equal to A or 3.

so its B.

good question....
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Re: DS problem solving [#permalink] New post 02 Apr 2008, 07:05
Yeah Gmat tiger, but what if ab was 31? that satisfies the condition. so it is insuff in that it tells you 2 different things.

E.

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Re: DS problem solving [#permalink] New post 02 Apr 2008, 07:53
how do you know AB and BA are not something like 11?

stem doesnt explicitly say that A and B are different..

GMAT TIGER wrote:
young_gun wrote:
AB and BA are 2-digits integers and AB*BA=CDE, where CDE is a 3-digits integer. What is AB?
1). A=E
2). A=C


we know AB and BA are two digit integers so 0 cannot be any of the digit.
also we know CDE is a 3 digit integer so it is < than 1000 but > than 121. so:-

i. digits can only be 1, 2, 3, 4, 5, and 6.
ii. 1000 > CDE > 121.

from 1: AB can be of any value from 11, 12, 13, 14, 15, 16, 21, 31 41, 51, and 61. any value of AB results in A=E if unit digit of AB is 1. so nsf.
from 2: again AB can be of any value from 11, 12, 13, 14, 15, 16, 21, 31 41, 51, and 61. but to have A=C, AB must be 21.

if AB is 13, ABxBA = 403. so C=4, which is not equal to A or 3.

so its B.

good question....
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Re: DS problem solving [#permalink] New post 02 Apr 2008, 10:48
GMAT TIGER wrote:
young_gun wrote:
AB and BA are 2-digits integers and AB*BA=CDE, where CDE is a 3-digits integer. What is AB?
1). A=E
2). A=C


we know AB and BA are two digit integers so 0 cannot be any of the digit.
also we know CDE is a 3 digit integer so it is < than 1000 but > than 121. so:-

i. digits can only be 1, 2, 3, 4, 5, and 6.
ii. 1000 > CDE > 121.

from 1: AB can be of any value from 11, 12, 13, 14, 15, 16, 21, 31 41, 51, and 61. any value of AB results in A=E if unit digit of AB is 1. so nsf.
from 2: again AB can be of any value from 11, 12, 13, 14, 15, 16, 21, 31 41, 51, and 61. but to have A=C, AB must be 21.

if AB is 13, ABxBA = 403. so C=4, which is not equal to A or 3.

so its B.

good question....


I overlooked for 11.

come down to E.
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Re: DS problem solving [#permalink] New post 03 Apr 2008, 03:40
11*11=121
21*12=252
31*13=403

from this we can easily deduce that ans is E

bcoz we need to figure out what is AB

so AB could be 11, 21, 31......................
Re: DS problem solving   [#permalink] 03 Apr 2008, 03:40
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