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# Which of the following is equal to 2^k*5^(k-1)?

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Manager
Joined: 11 Jan 2007
Posts: 196
Location: Bangkok
Which of the following is equal to 2^k*5^(k-1)? [#permalink]

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05 Jun 2007, 20:37
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99,999^2 - 1^2 =

A. 10^10 - 2
B. (10^5 – 2)^2
C. 10^4(10^5 – 2)
D. 10^5(10^4 – 2)
E. 10^5(10^5 – 2)

Which of the following is equal to (2^k)(5^(k − 1))?

A. 2(10^k − 1)
B. 5(10^k − 1)
C. 10^k
D. 2(10^k )
E. 10^(2k − 1)
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Joined: 26 Feb 2006
Posts: 899

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05 Jun 2007, 21:36
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jet1445 wrote:
99,999^2 - 1^2 =

A. 10^10 - 2
B. (10^5 – 2)^2
C. 10^4(10^5 – 2)
D. 10^5(10^4 – 2)
E. 10^5(10^5 – 2)

E.
= 99,999^2 - 1^2
= (10^5 - 1)^2 - 1
= (10^5)^2 - 2 (10^5) + 1 - 1
= 10^5 (10^5 - 2)
Director
Joined: 26 Feb 2006
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05 Jun 2007, 21:45
jet1445 wrote:
Q18: Which of the following is equal to (2^k)(5^(k − 1))?

A. 2(10^k − 1)
B. 5(10^k − 1)
C. 10^k
D. 2(10^k )
E. 10^(2k − 1)

= (2^k) {5^(k − 1)}
= (2^k) (5^k) / 5
= 10^k/5

none???????? seems the answeres have typo.
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Joined: 17 May 2007
Posts: 2950

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06 Jun 2007, 05:02
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Himalayan I did it that way too, but unfortunately to match the answer I had to redo it slightly differently

2^k * 5 ^(k-1)
= 2 * 2^(k-1) * 5 ^(k-1)
= 2 * 10^(k-1)

which is option A
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Re: 99,999^2 - 1^2 = A. 10^10 - 2 B. (10^5 2)^2 C. [#permalink]

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16 Feb 2012, 22:02
For second question Ans. should be (10^k)/5

Further help appreciated.
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Re: 99,999^2 - 1^2 = A. 10^10 - 2 B. (10^5 2)^2 C. [#permalink]

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16 Feb 2012, 22:18
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Baten80 wrote:
For second question Ans. should be (10^k)/5

Further help appreciated.

(10^k)/5 is a correct solution, though as it's not listed among the answer choices then we should manipulate with given expression in another way.

Which of the following is equal to 2^k*5^(k-1)?
A. 2*10^(k-1)
B. 5*10^(k-1)
C. 10^k
D. 2*10^k )
E. 10^(2k-1)

$$2^k*5^{k-1}=(2*2^{k-1})*5^{k-1}=2*10^{k-1}$$.

99,999^2 - 1^2 =
A. 10^10 - 2
B. (10^5 – 2)^2
C. 10^4(10^5 – 2)
D. 10^5(10^4 – 2)
E. 10^5(10^5 – 2)

Apply $$a^2-b^2=(a+b)(a-b)$$:

$$99,999^2-1^2 =(99,999+1)(99,999-1)=10^5(10^5-2)$$

Hope it's clear.
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Re: Which of the following is equal to 2^k*5^(k-1)? [#permalink]

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04 Jun 2012, 12:31
Bunuel,
Can you please explain how (2^k)*(5^(k-1)) equals (2*(2^k-1))*5^k-1

I have seen this logic on 2 other problems now and this one step is causing me to miss questions. If you could please explain you get that it would help me a lot!
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Joined: 02 Sep 2009
Posts: 39755
Re: Which of the following is equal to 2^k*5^(k-1)? [#permalink]

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04 Jun 2012, 12:46
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wjf9838 wrote:
Bunuel,
Can you please explain how (2^k)*(5^(k-1)) equals (2*(2^k-1))*5^k-1

I have seen this logic on 2 other problems now and this one step is causing me to miss questions. If you could please explain you get that it would help me a lot!

Operations involving the same bases:
Keep the base, add or subtract the exponent (add for multiplication, subtract for division)
$$a^n*a^m=a^{n+m}$$

$$\frac{a^n}{a^m}=a^{n-m}$$

So, $$2*2^{k-1}=2^1*2^{k-1}=2^{1+k-1}=2^k$$.

For more on exponents check Number Theory chapter of Math Book: math-number-theory-88376.html

Hope it helps.
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Posts: 16027
Re: Which of the following is equal to 2^k*5^(k-1)? [#permalink]

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06 Apr 2014, 07:34
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Re: Which of the following is equal to 2^k*5^(k-1)? [#permalink]

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15 May 2015, 07:57
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Which of the following is equal to 2^k*5^(k-1)?   [#permalink] 15 May 2015, 07:57
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