GCF - DS

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GCF - DS

by shankar.ashwin » Mon Nov 21, 2011 9:10 am
If the expression #(a,b)# is defined to be the sum of a and b divided by the greatest common factor of a and b, what is the value of #(6,x)# ?

A) GCF of 6 and x is 1/7 the sum of 6 and x.

B) x is a positive odd integer not equal to 1.

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by HSPA » Mon Nov 21, 2011 9:16 am
IMO A
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by neelgandham » Mon Nov 21, 2011 9:27 am
If the expression #(a,b)# is defined to be the sum of a and b divided by the greatest common factor of a and b, what is the value of #(6,x)# ?
A) GCF of 6 and x is 1/7 the sum of 6 and x.
#(6,x)# = (6+x)/((1/7)*(6+x)) = 7
Sufficient!
B) x is a positive odd integer not equal to 1.
Let the value of x be 3, then #(6,x)# = (6+3)/3 = 3
Let the value of x be 5, then #(6,x)# = (6+5)/1 = 11
Insufficient!

IMO A
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by pemdas » Mon Nov 21, 2011 9:23 pm
what's OA? this one is a bit tricky, i agree about A
#(a,b)#=(a+b)/GCF, find #(6,x)#?
notes: the expression can be defined for all x integers not equal to 0 (x can be positive and negative),LCM*GCF=ab=6x

st(1) GCF=(6+x)/7, #(6,x)#=7(6+x)/(6+x)
x=8, GCF=2 and #(6,x)#=(6+8)/2=7

x=-20, GCF=2
By definition: "The greatest common divisor (gcd), also known as the greatest common factor (gcf), or highest common factor (hcf), of two or more non-zero integers, is the largest positive integer that divides the numbers without a remainder." Hence only GCF=2 counts here (not -2), #(6,x)#=(6-20)/2=-7 ???
Considering the condition GCF=(6+x)/7 must be positive, x is positive, Sufficient (ans=7)

st(2) x>1 (x is odd) -> (6+x)/GCF(6,odd x) -> x can be 3 or 5 Not Sufficient
shankar.ashwin wrote:If the expression #(a,b)# is defined to be the sum of a and b divided by the greatest common factor of a and b, what is the value of #(6,x)# ?

A) GCF of 6 and x is 1/7 the sum of 6 and x.

B) x is a positive odd integer not equal to 1.
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