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Re: If x is divisible by 18 and y is divisible by 12, which of t [#permalink]
So I am right. You've made a mistake posting answer as C.
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Re: If x is divisible by 18 and y is divisible by 12, which of t [#permalink]
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Simon wrote:
So I am right. You've made a mistake posting answer as C.

Yup! sorry about that it is A.
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Re: If x is divisible by 18 and y is divisible by 12, which of t [#permalink]
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First number 18a, 2nd 12b

x+y=18a+12b=6*(3a+2b) hence multiple of 6

xy=18a*12b=8*27*ab
Now a and b both could be 1 so min 8*27 is not multiple of 48

x/y=18a/12b=3a/2b
a & b could be 1
So x/y=3/2 not multiple of 6
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If x is divisible by 18 and y is divisible by 12, which of t [#permalink]
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The prime factors of 18 = 3, 3, 2

The prime factors of 12 = 2, 2, 3

A)

x+y is divisible by 6.

Now x+y should at least contain 18+12=30 which is divisible by 6. Hence x+y is divisible by 6. As all other values of x and y would be multiples of 18 and 12 and would contain at least 30 inside them.

B) xy is divisible by 48

The factors of xy are the factors of x and y taken together = 3, 3, 2, 2, 2, 3

Now you cannot create 48 by multiplying the prime factors of xy listed above

So xy is NOT divisible by 48

C)

\(\frac{x}{y}\) is divisible by \(6\)

\(\frac{x}{y} = \frac{18}{12} = \frac{3}{2}\). Hence it is NOT divisible by \(6\).

The answer is only A
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If x is divisible by 18 and y is divisible by 12, which of t [#permalink]
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