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# X = 2^a * 3^b, while Y = 3^c * 5^d, where a, b, c, d are all positive

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X = 2^a * 3^b, while Y = 3^c * 5^d, where a, b, c, d are all positive  [#permalink]

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25 Apr 2018, 11:02
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55% (hard)

Question Stats:

58% (02:14) correct 42% (01:55) wrong based on 57 sessions

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X = 2^a * 3^b, while Y = 3^c * 5^d, where a, b, c, d are all positive integers. What is the Greatest Common Divisor of X and Y?

(1) Lowest Common Multiple of X and Y is (2^3 * 3^2 * 5^4).

(2) a, b, c, d are all distinct from each other.
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X = 2^a * 3^b, while Y = 3^c * 5^d, where a, b, c, d are all positive  [#permalink]

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25 Apr 2018, 11:53
amanvermagmat wrote:
X = 2^a * 3^b, while Y = 3^c * 5^d, where a, b, c, d are all positive integers. What is the Greatest Common Divisor of X and Y?

(1) Lowest Common Multiple of X and Y is (2^3 * 3^2 * 5^4).

(2) a, b, c, d are all distinct from each other.

as between X & Y only common prime factor is 3, hence to find the GCD, we need to know the powers of 3 i.e. the value of $$b$$ & $$c$$

Statement 1: from this we know that one of the powers of 3 is 2 but other power could be 2, or 1. Hence the GCD could be $$3^2$$, or $$3$$. Insufficient

Statement 2: no values can be found out from this statement. Insufficient

Combining 1 & 2: one of the power of 3 is 2 hence other power of 3 has to be 1 as b & c are distinct integers. Hence the GCD will be $$3$$. Sufficient

Option C
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Re: X = 2^a * 3^b, while Y = 3^c * 5^d, where a, b, c, d are all positive  [#permalink]

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13 May 2018, 09:20
amanvermagmat wrote:
X = 2^a * 3^b, while Y = 3^c * 5^d, where a, b, c, d are all positive integers. What is the Greatest Common Divisor of X and Y?

(1) Lowest Common Multiple of X and Y is (2^3 * 3^2 * 5^4).

(2) a, b, c, d are all distinct from each other.

1): we know that one of the powers of 3 is 2 but other power could be 2, or 1. Hence the GCD could be 3 or 3^2.

3 to power zero cannot be considered as a,b,c,d are positive and zero is neither positive nor negative.

Insufficient

2) no values can be found out from this statement. Insufficient

Combining: one of the power of 3 is 2 hence other power of 3 has to be 1 as b & c are distinct integers.

Note: zero cannot be considered in combining values as zero is not positive integer

Hence the GCD will be 3. Sufficient

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Re: X = 2^a * 3^b, while Y = 3^c * 5^d, where a, b, c, d are all positive  [#permalink]

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14 May 2018, 01:03
amanvermagmat wrote:
X = 2^a * 3^b, while Y = 3^c * 5^d, where a, b, c, d are all positive integers. What is the Greatest Common Divisor of X and Y?

(1) Lowest Common Multiple of X and Y is (2^3 * 3^2 * 5^4).

(2) a, b, c, d are all distinct from each other.

From stem: Definition of GCD: GCD is the product of common primes in their least power.
The only prime factor common to X and Y is 3. So, the GCD of X and Y will be 3^b is b < c and the GCD will be 3^c if c < b.

Statement 1: LCM of X and Y = (2^3 * 3^2 * 5^4)
Definition of LCM: LCM is the product of all primes in their highest power.
Relevant information: The highest power of 3 between X and Y is 3^2 as the LCM contains 3^2.
Possibility 1: The two powers of 3 found in X and Y could both be 3^2. In that case, the GCD will be 3^2.
Possibility 2: The two powers of 3 found in X and Y could be 3^1 and 3^2. In that case, the GCD will be 3^1.

With the information in statement 1, we will not be able determine whether it is 3^2 or 3^1.
Statement 1 ALONE is NOT sufficient.

Statement 2:a, b, c, d are all distinct from each other.
It leaves the door open to infinite possibilities.
Statement 2 ALONE is NOT sufficient.

Combining the 2 statements: From statement 2, if a, b, c, and d are distinct, b and c will have to be different numbers.
From statement 1, we narrowed down values of b and c to be both 2 or one 1 and the other 2.
If b and c are distinct, both cannot be same. So, that rules of possibility 1 of statement 1.
We are left with only possibility 2 that satisfies both conditions. So, the GCD is 3^1.

TOGETHER staements SUFFICIENT.
Choice C.
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Re: X = 2^a * 3^b, while Y = 3^c * 5^d, where a, b, c, d are all positive   [#permalink] 14 May 2018, 01:03
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