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# Z=2a×5b×7c A positive integer Z can be expressed in terms of its prime

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Manager
Joined: 14 Sep 2015
Posts: 65
Location: India
GMAT 1: 700 Q45 V40
GPA: 3.41
Z=2a×5b×7c A positive integer Z can be expressed in terms of its prime [#permalink]

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23 May 2017, 00:34
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Difficulty:

95% (hard)

Question Stats:

37% (02:10) correct 63% (02:11) wrong based on 52 sessions

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$$Z=2^a×5^b×7^c$$

A positive integer Z can be expressed in terms of its prime factors as above, where a, b and c are positive integers. Is $$3^{|a−b|}<9$$?

(1) Z is divisible by 40 but not by 50
(2) $$Z^{a−b}=2^6×5^2×7^4$$
Math Expert
Joined: 02 Sep 2009
Posts: 46250
Z=2a×5b×7c A positive integer Z can be expressed in terms of its prime [#permalink]

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23 May 2017, 00:50
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1
$$Z=2^a×5^b×7^c$$

A positive integer Z can be expressed in terms of its prime factors as above, where a, b and c are positive integers. Is $$3^{|a−b|}<9$$?

Is $$3^{|a−b|}<9$$?

Is $$3^{|a−b|}<3^2$$?

Is $$|a−b|<2$$?

(1) Z is divisible by 40 but not by 50

$$40 = 2^3*5$$ is a factor of Z while $$50 = 2*5^2$$ is not. This implies that a (the power of 2) is at least 3 and b (the power of 5) is exactly 1: $$a > 3$$ and $$b = 1$$. Thus, $$|a−b|$$ will be greater than 2. Sufficient.

(2) $$Z^{a−b}=2^6×5^2×7^4$$.

$$Z^{a−b}=2^6×5^2×7^4=(2^a×5^b×7^c)^{a-b}$$.

$$2^6×5^2×7^4=(2^a×5^b×7^c)^{a-b}$$.

$$2^6×5^2×7^4=2^{a(a-b)}×5^{b(a-b)}×7^{c(a-b)}$$.

$$6 = a(a-b)$$ and $$2 = b(a-b)$$

Divide one by another $$3 = \frac{a}{b}$$ --> $$a = 3b$$.

$$|a−b|=|2b|$$. Since b is a positive integer, then |2b| is at least 2. Thus, $$|a−b|=|2b|$$ will be greater than 2. Sufficient.

Hope it'clear.
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Joined: 23 Feb 2017
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Re: Z=2a×5b×7c A positive integer Z can be expressed in terms of its prime [#permalink]

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23 May 2017, 15:59
Bunuel, 40 = 2^3*5 and 50 = 2*5^2.
I agree that a is atleast 3.. but how is b EXACTLY 1.. I can do Z= 200, which is 2^3*5^2*7^0 .. so b can be 2 as well right..
am I missing something?
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Joined: 02 Sep 2009
Posts: 46250
Re: Z=2a×5b×7c A positive integer Z can be expressed in terms of its prime [#permalink]

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23 May 2017, 21:40
sasidharrs wrote:
Bunuel, 40 = 2^3*5 and 50 = 2*5^2.
I agree that a is atleast 3.. but how is b EXACTLY 1.. I can do Z= 200, which is 2^3*5^2*7^0 .. so b can be 2 as well right..
am I missing something?

We are told that z is divisible by 5 but not by 5^2, so the power of 5 in z must be 1.
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Re: Z=2a×5b×7c A positive integer Z can be expressed in terms of its prime   [#permalink] 23 May 2017, 21:40
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