If x and y are positive integers, which of the following

Hi All,

This question can be solved by TESTing VALUES. Here though, we'll be TESTing VALUES to prove which answers are POSSIBLE (so that we can eliminate them). This prompt is actually built around a few Number Properties and we can actually use the answer choices to our advantage (instead of just randomly TESTing VALUES until we knock out four of the answers).

We're told that X and Y are POSITIVE INTEGERS. We're asked which of the 5 answers CANNOT be the GCD of 35X and 20Y. This means that 4 of the answers COULD be the GCD under certain circumstances.

Before we TEST anything, there are some things to note about 35X and 20Y:

1) 20Y will ALWAYS be EVEN (20, 40, 60, 80, 100, etc.)
2) 35X can be EVEN or ODD (35, 70, 105, 140, etc.)

Looking at the list of answer choices, there are some that we can quickly eliminate...

IF
X = 1
Y = 1
Then we have 35 and 20, so the GCD = 5
Eliminate A.

X = 20
Y = 1
Then we have 700 and 20, so the GCD = 20
Eliminate D.

X = 1
Y = 35
Then we have 35 and 700, so the GCD = 35
Eliminate E.

With the remaining 2 answers, we have to think a little more.

To get a simple example of 5(X-Y) to be the GCD, we probably need 35X to be ODD

IF...
X = 3
Y = 2
Then we have 105 and 40, so the GCD = 5
5(X-Y) = 5(3-2) = 5
This can ALSO be the GCD
Eliminate B.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Number divided by HCF should be an integer.

35x divided by 20x is not an integer hence 20x cannot be the HCF.

Theory: GCD/HCF of two numbers is the highest/greatest factor which is common between the two numbers.

So, GCD of two numbers should always divide the two numbers:--

A. 5 : 5 divides both 35x and 20y so 5 can be the GCD

B. 5(x-y) : By taking values it can be proved that 5(x-y) can be GCD (take x=2 and y =1 then 5(x-y) = 5(2-1) = 5 which divides both 35x and 20y

C. 20x : For any positive integer value of x, 20x can never divide 35x (because 35x/20x = 7/4). Since, 20x is not a factor of 35x so it CANNOT be the GCD of 35x and 20y

D. 20y : GCD of two numbers is always less than or equal to the smallest of the two numbers. By taking numbers it can be proved that 20y can be GCD (take values of x and y such that 35x = some integer * 20y Example: y = 1 and x = 4)

E. 35x : Same as Option d. Take values of x and y such that 20y = 35x * integer Example: x = 1 and y = 7)

So, Answer will be C
Hope it helps!

To Learn about the Basics of LCM and GCD watch the following video

Hi paragpednekar,

This question can be solved with math "theory" or by TESTing VALUES. Here's how to eliminate the 4 wrong answers by TESTing VALUES...

We're told that X and Y are POSITIVE INTEGERS. We're asked which of the following CANNOT be the greatest common divisor of 35x and 20y.

Answer A: 5
IF...X = 1, Y = 1...
35 and 20 have a GCD of 5.
Eliminate Answer A

Answer B: 5(X-Y)
IF...X = 3, Y = 2...
105 and 40 have a GCD of 5. 5(3-2) = 5
Eliminate Answer B

Answer D: 20Y
IF...X = 4, Y = 1...
140 and 20 have a GCD of 20. 20(1) = 2
Eliminate Answer D

Answer E: 35X
IF...X = 2, Y = 7...
70 and 140 have a GCD of 70. 35(2) = 70
Eliminate Answer E

There's only one answer left....

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Hi All,

This question can be solved with math "theory" or by TESTing VALUES. Here's how to eliminate the 4 wrong answers by TESTing VALUES...

We're told that X and Y are POSITIVE INTEGERS. We're asked which of the following CANNOT be the greatest common divisor of 35x and 20y.

Answer A: 5
IF...X = 1, Y = 1...
35 and 20 have a GCD of 5.
Eliminate Answer A

Answer B: 5(X-Y)
IF...X = 3, Y = 2...
105 and 40 have a GCD of 5. 5(3-2) = 5
Eliminate Answer B

Answer D: 20Y
IF...X = 4, Y = 1...
140 and 20 have a GCD of 20. 20(1) = 2
Eliminate Answer D

Answer E: 35X
IF...X = 2, Y = 7...
70 and 140 have a GCD of 70. 35(2) = 70
Eliminate Answer E

There's only one answer left....

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Please find the video solution oft eh question as attached here. Subscribe our youtube channel if you want more such concepts and video

The video solution is as available here.



Answer: Option C

Lets split the numbers into factors and see where we can see a fit for the answer choices.
\(35x:\) \(2^0\) \(5^1\) \(7^1\) \(x\)
\(20y:\) \(2^2\) \(5^1\) \(7^0\) \(y\)

A): As you can see, 5 is already the GCF of 20 and 35. So 20y and 35x have at least 5 as GCF.
\(2^0\) \(5^1\) \(7^1\) \(x\)
\(2^2\) \(5^1\) \(7^0\) \(y\)
Incorrect.

B): As discussed in A), 5 is already part of the GCF. We only have to look whether we can bring in (x-y) into both numbers. And yes, indeed, thats possible. If x=2y, then (x-y)=(2y-y)=y is included in both, x and y!
\(2^0\) \(5^1\) \(7^1\) \((x=2y)\)
\(2^2\) \(5^1\) \(7^0\) \(y\)
Incorrect.

C): Well, we would need 20 and x as factors in both of them(35x and 20y). 35x lacks the 20, so x has to bring two 2s. But then, for x also to be included in the GCF, y would have to include x (which is possible) and x would have to include itself, in addition to two 2s, because x also has to bring in those two 2s.. So paradoxically, x needs to be at least 20x, which is not possible.
\(2^0\) \(5^1\) \(7^1\) \((x=20x !?)\)
\(2^2\) \(5^1\) \(7^0\) \((y=x)\)
Correct.

D): 20y requires two 2s and one 5 and one y. If all of them would be included in 35x, we would be able to call at least 20y the GCF. So lets make 35x include 20y. All we need for that is two 2s and y. If we include at least them in x, then at least 20y is the GCF.
\(2^0\) \(5^1\) \(7^1\) \((x=2^2y)\)
\(2^2\) \(5^1\) \(7^0\) \(y\)
Incorrect.

E): We can play the same game as in D. For 35x to be the GCF, let 35x be included in 20y. Therefore, y needs to have at least 7x.
\(2^0\) \(5^1\) \(7^1\) \(x\)
\(2^2\) \(5^1\) \(7^0\) \((y=7x)\)
Incorrect.

One of my students asked me to solve this question. Please forgive me if my solution looks like how does solutions.

Key concept: If k is a divisor of n, then n/k is an INTEGER
For example, since 8 is a divisor of 24, it is also true that 24/8 is an integer
Likewise, since 14 is a divisor of 14, it is also true that 14/14 is an integer

The question asks, "Which of the following CANNOT be the greatest common divisor of 35x and 20y?"

C) If 20x is the greatest common divisor of 35x and 20y, then it must be true that 35x/20x and 20y/20x are INTEGERS
Notice that 20y/20x COULD be an integer if x is a divisor of y.
However, 35x/20x simplifies to be 7/4, which is definitely NOT an integer (regardless of the value of x)
As such, 20x can never be the greatest common divisor of 35x and 20y

Cheers,
Brent

Given

• 35x and 20y are two numbers such that x and y are integers.

To Find

• The option that cannot be the greatest common divisor of 35x and 20y.

Approach and Working Out

• The numbers are 35x and 20y.

o The numbers can be written as (5 × 7x) and (5 × 4y)

• So, we need to use the terms 7x and 4y.

• Option A. possible since 7x and 4y can be co-primes.
• Option B. possible since when x – y = 1 such as 3 and 2.
• Option D, possible when x = 4y.
• Option E, possible when y = 7x.
• Option C, not possible.

Correct Answer: Option C

Solution:

Since 20x is not a factor of 35x, regardless of what x is, 20x can’t be the greatest common divisor of 35x and 20y.

Answer: C

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