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AB and BA are 2digits integers and AB*BA=CDE, where CDE is [#permalink]
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01 Apr 2008, 17:43
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AB and BA are 2digits integers and AB*BA=CDE, where CDE is a 3digits integer. What is AB? 1). A=E 2). A=C



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Re: DS problem solving [#permalink]
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01 Apr 2008, 19:20
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]
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01 Apr 2008, 19:32
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]
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01 Apr 2008, 20:08
young_gun wrote: AB and BA are 2digits integers and AB*BA=CDE, where CDE is a 3digits 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]
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01 Apr 2008, 23:03
young_gun wrote: AB and BA are 2digits integers and AB*BA=CDE, where CDE is a 3digits 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]
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02 Apr 2008, 08:05
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.
OA?



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Re: DS problem solving [#permalink]
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02 Apr 2008, 08:53
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 2digits integers and AB*BA=CDE, where CDE is a 3digits 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]
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02 Apr 2008, 11:48
GMAT TIGER wrote: young_gun wrote: AB and BA are 2digits integers and AB*BA=CDE, where CDE is a 3digits 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]
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03 Apr 2008, 04:40
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
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