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can someone tell me the shortcut for this silly problem?

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Intern
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can someone tell me the shortcut for this silly problem? [#permalink] New post 04 Jul 2007, 14:19
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can someone tell me the shortcut for this silly problem?

(5^21) x (4^11) = (2 x 10^n) , what is the value of n?

thanks :)
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 [#permalink] New post 04 Jul 2007, 15:11
no shortcut, but may be it helps?

5^11*5^10*4^11=2*10^n
(5*4)^11 * 5^10 =
20^10*20*5^10 =
100^10 * 20 = 2 * 10^n | :2
10^10*10^10 * 10 = 10^n
10^21 = 10^n

cya
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 [#permalink] New post 04 Jul 2007, 17:09
I would do it a slightyly different way:

(5^21)*(4^11)=2*10^n
(5^21)*(2^22)=2*10^n
(5^21)*(2)(2^21)=2*10^n
2*(5^21)*(2^21)=2*10^n
2*10^21...
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 [#permalink] New post 04 Jul 2007, 17:45
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5^21 * 2^22 = 5^21 * 2^21 * 2 = 10^21 * 2

so n = 21
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Re: (5^21) x (4^11) = (2 x 10^n) [#permalink] New post 04 Jul 2007, 23:45
dzelkas wrote:
can someone tell me the shortcut for this silly problem?

(5^21) x (4^11) = (2 x 10^n) , what is the value of n?

thanks :)


(5^21)(2^22)=2*(2*5)^n
(5^21)(2^22)=2*2^n*5^n
(5^21)(2^22)=2^n+1*5*n
Equate either term to solve for n

5^21=5^n => n=21
or
2^22=2^n+1 => n=21
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Re: (5^21) x (4^11) = (2 x 10^n) [#permalink] New post 05 Jul 2007, 20:55
dzelkas wrote:
can someone tell me the shortcut for this silly problem?

(5^21) x (4^11) = (2 x 10^n) , what is the value of n?

thanks :)


Start with this fact: 4 = 2^2

From the expression: 5^21 * (2^2)^11 = 5^21 * 2^22

Now, include this fact: 10 = 5*2

(5*2)^21 * 2 = 10^21 * 2

Therefore, n = ???

Yep, 21.
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 [#permalink] New post 05 Jul 2007, 21:24
Shortcut:

Since you know that there is a 5^21 on the left side. You know that for the equation to be equal, there needs to be exactly 21 5's on the right side. The only way to create 21 5's is to have n=21.
  [#permalink] 05 Jul 2007, 21:24
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can someone tell me the shortcut for this silly problem?

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