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wat is c+w-x

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wat is c+w-x

by gmatnmein2010 » Mon Feb 08, 2010 5:39 am
Multiplication of the two digit numbers wx and cx, where w,x and c are unique non-zero digits, the product is a three digit number. What is w+c-x?
(1) The three digits of the product are all the same and different from w c and x.
(2) x and w+c are odd numbers.
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by harsh.champ » Mon Feb 08, 2010 5:51 am
gmatnmein2010 wrote:Multiplication of the two digit numbers wx and cx, where w,x and c are unique non-zero digits, the product is a three digit number. What is w+c-x?
(1) The three digits of the product are all the same and different from w c and x.
(2) x and w+c are odd numbers.
I wouild go with C.
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by thephoenix » Mon Feb 08, 2010 10:21 am
IMO A

as per q + s1)
wx * Cx= nnn where the product is 3 digit same no. and diff from w,c,x

now the unit dig in the multiplication is x^2 and hence the unit dig of reqd value cant be 2,3,5 or 7 ; 5 bcoz then x must be a 5 and in that case x and n will be same , which violates the condition

hence it can be 111,444.666 or 999

now
for 111 : it has only 2 factors 37*3 but in that the other no is not a two dig one hence ruled out
for 444: its factors are 2*2*3*37 and hence it must be 37*12 but in that the unit dig of bth the no's differ hence ruled out
for 666: its factors are 3*2*3*37=37*19 but again in that the unit dig of bth the no's differ hence ruled out
so only left is 999 its factors are 3*3*3*37=37*27--->satisfies all the condition
w+c_x--->3+2-7=-1
hence suff

now asper q +s2)
x can be anything 1,3,5,7,9(odd no's)
and w+c = 1,3,5,9(any odd sum)
when w+c=1 for all x we have diff ans
hence insuff
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by ajith » Tue Feb 09, 2010 3:10 am
gmatnmein2010 wrote:Multiplication of the two digit numbers wx and cx, where w,x and c are unique non-zero digits, the product is a three digit number. What is w+c-x?
(1) The three digits of the product are all the same and different from w c and x.
(2) x and w+c are odd numbers.
(10w+x)(10c+x) = 100wc + 10 (c+w) + x^2
wc<10

1.) the product is a multiple of 111 = 37*3*k
the only possibility is 37*27 = 999 (the only other possibility was 74*12 which violates conditions)
Sufficient

2) Insufficient
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by arora007 » Thu Aug 12, 2010 6:15 am
wow this one was nice...

1 ) says 3 digits are same... so it could be 111*(between 1 & 9)

111=37*3

we need some multiple of 3 which is 2 digited and ends in 7.
3*9 =27

so 3+7+2 = 12
SUFFICIENT....

2) 19 * 29

1+2 =3 is odd
9 is odd

so 1+2+9 = 12

29 * 39

2+3= 5 odd
9 is odd

2+3+9 = 14

thus the solution is not unique in sufficient...

ans A
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