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# If a, b, and c are positive integers, and a is a multiple of 3

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CrackVerbal Quant Expert
Joined: 12 Apr 2019
Posts: 113
If a, b, and c are positive integers, and a is a multiple of 3  [#permalink]

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26 May 2019, 01:17
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If a, b, and c are positive integers, and a is a multiple of 3, do c and 18 have a common factor greater than 1?

I. a/9 +b/6 = c/18
II.a/6 +b/9 = c/18
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Re: If a, b, and c are positive integers, and a is a multiple of 3  [#permalink]

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26 May 2019, 19:16
ArvindCrackVerbal wrote:
If a, b, and c are positive integers, and a is a multiple of 3, do c and 18 have a common factor greater than 1?

I. a/9 +b/6 = c/18
II.a/6 +b/9 = c/18

a, b and c>0 (I)
a=3x, x is I

we have to find common factor (c,18) >1
Statement 1:

cross multiply-
2a + 3b = c
6x + 3b = c
3 (2x + 1) = c, so C is a multiple of 3. Therefore common factor (c,18) definitely 3.

Statement 2:
cross multiply-
3a + 2b = c
9x + 2b = c, now c plugging values we can see that c can be multiple of 3 as well as 11. Insufficient.

IMO only A
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CrackVerbal Quant Expert
Joined: 12 Apr 2019
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Re: If a, b, and c are positive integers, and a is a multiple of 3  [#permalink]

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29 May 2019, 20:25
Any pair of numbers which have only 1 as a common factor are called Co-prime numbers. If c and 18 have a common factor greater than 1, they will not be co-prime. So, essentially, we are trying to find out if c is co-prime with 18. Therefore, we will have to work on finding out an expression/equation to find c.

Using Statement I alone, we know, $$\frac{a}{9}$$ + $$\frac{b}{6}$$ = $$\frac{c}{18}$$. On simplifying the above, we can say,
2a + 3b = c.

2a will definitely be a multiple of 3 since a is a multiple of 3. 3b will definitely be a multiple of 3. So, c will be a multiple of 3. If c is a multiple of 3, c and 18 have a common factor of 3 and so, have a common factor greater than 1.
Statement I alone is sufficient to answer the question with a definite YES. So, the possible answers are A or D. Options B, C and E can be eliminated.

Using statement II alone, we know, $$\frac{a}{6}$$ + $$\frac{b}{9}$$ = $$\frac{c}{18}$$. On simplifying the above, we get,
3a + 2b = c.

From this, we can only deduce that 3a is a multiple of 3. Since we do not know the nature of b, we cannot say anything about 2b. Therefore, we cannot say conclusively whether c will be a multiple of 3. Hence, we cannot deduce if c and 18 will have a common factor greater than 1.
Statement II alone is insufficient. Option D can be eliminated. So, the correct answer option is A.

It’s a good idea to always analyse the question stem as much as you can, rather than trying values. Practicing the analysis method will also help you to consolidate concepts, because, when analyzing a question stem, your endeavor will always be to look for what concept/s can be applied in the given situation.

Hope this helps!
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If a, b, and c are positive integers, and a is a multiple of 3  [#permalink]

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29 May 2019, 20:51
ArvindCrackVerbal wrote:
If a, b, and c are positive integers, and a is a multiple of 3, do c and 18 have a common factor greater than 1?

I. a/9 +b/6 = c/18
II.a/6 +b/9 = c/18

Consider the first equation,

a/9 + b/6 = c/18
Simflied equation will be as below

a/3 + b/2 = c/6

Here "a/3" will be a integer."b/2" will be a integer if b is even.

CaseI: b is even

Integer = C/6, So 'c' must be a multiple of 6, to satisfy this equation.which implies C must be a multiple of 18.

CaseII: b is odd

Integer + b/2 = c/6

It will become like

Integer/2 = c/6, from this C must be multiple of 3 which implies c must be a multiple of 18.

So, A is sufficient.

Now we check for B.

a/6 + b/9 = c/18

Simplified equation will be

a/2 + b/3 = c/6

Now 'a' can even or odd multiple of 3.

CaseI : a is even

Then, integer + b/3 = c/6

So the equation can be integer/3 = c/6

From this C must be even , which implies it is multiple of 18.

CaseII
if a is odd multiple of 3

a/2 + b/3 = c/6

So it will become like
Integer/6 = c/6
So c can any value from 5 to infinity. It is uncertain here.

B is out.

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If a, b, and c are positive integers, and a is a multiple of 3   [#permalink] 29 May 2019, 20:51
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