If a, b, c, and d are all positive integers, is c divisible by a/d?

For me it is C.

Using the first statement to define an equation: CB/A=k

Using the second statement to say that you can factorize by D.

With the both answers you can definitely answer to the question.

Can we have OA?

I'm going with E.

Based on part I, we know that a/d is an integer. THat means that d/a might be an integer, or it might not (if it's an integer, it would have to be 1, if d=a).

Based on part II, we know that C/A is an integer.. so to prove that DC/A is an integer, we would need proof, one way or the other, that D/A is an integer. Unfortunately, we can't make that distinction either positively or negatively for reasons discussed in the previous paragraph.

Firstly, I thought E

given a,b,c,d are positive integers. Is c divisible by a/d implies \(\frac{c}{a/d}=\frac{cd}{a}=int\)

(1) b * c is divisible by a
Factor 'a' can be present in 'b' or in 'c'..... We cannot exactly say if 'c' has factor 'a' to make \(\frac{cd}{a}\) integer because 'b' can also have a factor 'a' or both 'b', 'c' can have a factor of 'a'- Not Sufficient

(2) GCF (a,b) = d
GCF of a,b is d in the sense..... both 'a' and 'b' will have a common factor 'd'.... but you do not know exactly what is 'a'... 'a' can just be 'd' or '2d' or '3d' or 'nd' in which case \(\frac{cd}{a}\) has a remaining factor from 'a' (except 'a'='d' case all other cases have factors such as 2,3, or n) in the denominator..... For \(\frac{cd}{a}\) to be an integer the remaining factor of 'a' should also be present in 'c'. Not sufficient.

(1) + (2)
there are still possible cases of both things to happen...
My working was as below.
GCF (a,b) = d. In other words we can write as \(a=Ad\) and \(b=Bd\) where (A, B) should be (even, odd) or (odd, even) or (prime numbers).
From 1, \(c=integer*a/b\) implies \(c = integer*A/B\)

Now consider \(\frac{cd}{a}\). substituting values \(\frac{cd}{a}\)= \(integer*A/B *d/Ad\) = \(integer/B\). Since there are no restrictions with Integer and B, I went with E.... that fraction can be an integer or a non-integer.
But I am not able to create examples for showing \(\frac{cd}{a}\) is not integer are yielding contradictory values.... Answer may be C.... Can someone decode the logic from where I left off. Thanks.

You can also think logically without getting into a ton of variables.

Using both together:

GCF (a,b) = d means a and b both have d as a factor and nothing else common e.g. a = 2d, b = 3d etc
b*c is divisible by a. b and a have only d common so the leftover factors of a must be in c. That is, the 2 of a must be a factor of c so that b*c is divisible by a.

Question is whether c is divisible by a/d. a already has d as a factor and another factor (2 in our example). We have already established above that the leftover factor must be a factor of c i.e. 2 must be in c. Hence c is divisible by a/d (a/d is 2 in our example).

Answer (C)

Thank you for stopping by. I was the person who pm'd you. I messed it up above with too many variables. Your solution is now clear. Thank you.

Is this wrong ?
Ler b= 2 c= 3 a= 2

Then b*c is divisible by a
and GCD (a,b)= 2 hence d = 2

a/d = 2/2= 1
c= 3
Is 1 divisible by 3 ? No

Let b= 2 c= 3 a= 3
Then b*c is divisible by a
and GCD (a,b)= 1= d

a/d= 3/1 = 3 ,
C= 3
Is 3 divisible by 3? yes

How come I am getting E?

I recommend you to read the question stem once again. It says "is c divisible by a/d?" but in your first case of number plugging you tried the other way i.e. is a/d divisible by c? In your example, the statement should be is 3 divisible by 1? Yes.

Hi there!
From statement 2
Why p and q are coprime integers?