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If x= 1/(2^2*3^2*4^2*5^2) is expressed as a decimal, how many distinct

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If x= 1/(2^2*3^2*4^2*5^2) is expressed as a decimal, how many distinct  [#permalink]

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New post Updated on: 22 Jun 2015, 02:47
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If x= 1/(2^2*3^2*4^2*5^2) is expressed as a decimal, how many distinct nonzero digits will x have?

A. One
B. Two
C. Three
D. Seven
E. Ten

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Originally posted by sony1000 on 22 Jun 2015, 01:02.
Last edited by Bunuel on 22 Jun 2015, 02:47, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: If x= 1/(2^2*3^2*4^2*5^2) is expressed as a decimal, how many distinct  [#permalink]

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New post 22 Jun 2015, 01:41
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You don't really need to do long division here if you don't want to, as long as you know that 1/9 = 0.11111....., and 2/9 = 0.22222...., and so on.

If we multiply a number by 10, or by 100, or by any other power of 10, we're just moving the decimal point over. We won't change the number of distinct nonzero digits we have in our number. So we can freely multiply the fraction by any power of 10 without changing the answer. We can start by multiplying the fraction by 10^2 = 2^2 * 5^2, to get:

1/3^2*4^2

which is already a bit simpler. But we can keep going - we can multiply by 100^2 = (4*25)^2 = 4^2 * 25^2 to get the 4s out of the denominator as well. So this fraction

25^2/3^2 = 625/9

has the same number of distinct digits as the fraction in the question. Since 630/9 = 70, then 621/9 = 69, so 625/9 = 69 + 4/9 = 69.44444..... which has three distinct digits.

But I think most people would just do long division, either with 1/144, or with 1/14,400, which will also work, and probably is just as fast. I hate doing long division so I always look for another way!
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Re: If x= 1/(2^2*3^2*4^2*5^2) is expressed as a decimal, how many distinct  [#permalink]

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New post 22 Jun 2015, 02:55
sony1000 wrote:
If x= 1/(2^2*3^2*4^2*5^2) is expressed as a decimal, how many distinct nonzero digits will x have?

A. One
B. Two
C. Three
D. Seven
E. Ten


Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to rules 3, 7 and 10. Thank you.



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Hope it helps.
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Re: If x= 1/(2^2*3^2*4^2*5^2) is expressed as a decimal, how many distinct  [#permalink]

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New post 22 Jun 2015, 22:32
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sony1000 wrote:
If x= 1/(2^2*3^2*4^2*5^2) is expressed as a decimal, how many distinct nonzero digits will x have?

A. One
B. Two
C. Three
D. Seven
E. Ten


\(\frac{1}{(2^2*3^2*4^2*5^2)}\) = \(\frac{1}{2^2} * \frac{1}{3^2} * \frac{1}{4^2} * \frac{1}{5^2}= \frac{1}{100} * \frac{1}{9} * \frac{1}{16}\)

= \((.01 * .0625) * \frac{1}{9}\) (division by 9 leads to a non-terminating decimal so leave it for the end)

= .000625/9

= .0000694444...

So we have 3 distinct non zero digits.

Answer (C)
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Re: If x= 1/(2^2*3^2*4^2*5^2) is expressed as a decimal, how many distinct  [#permalink]

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New post 02 Oct 2017, 09:32
sony1000 wrote:
If x= 1/(2^2*3^2*4^2*5^2) is expressed as a decimal, how many distinct nonzero digits will x have?

A. One
B. Two
C. Three
D. Seven
E. Ten


we try to make dominator the multiple of 10,
so we multiple with 5^4

we have 5^4/9, we do not care how many 10 factors are there is the dominator

divide
694444444

so, there is only 3 distict

tricky
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Re: If x= 1/(2^2*3^2*4^2*5^2) is expressed as a decimal, how many distinct  [#permalink]

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New post 08 Aug 2018, 22:16
sony1000 wrote:
If x= 1/(2^2*3^2*4^2*5^2) is expressed as a decimal, how many distinct nonzero digits will x have?

A. One
B. Two
C. Three
D. Seven
E. Ten


Following is the simple calculation.

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If x= 1/(2^2*3^2*4^2*5^2) is expressed as a decimal, how many distinct  [#permalink]

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New post 25 Aug 2018, 14:10
Took 01:02.

1st) do some fast math in your head: 4*9*16*25 = 64 * 225 = 10* 225 + 50* 225 + 4 * 225 = 2250 + 50*200 + 50*25 + 900 = 12250 + 1250 + 900 = 14 400 (or: 100*16*10 - 1600 = 16000 - 1600 = 14400)
2nd) switch the fraction into a decimal
done

the easy way.
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Re: If x= 1/(2^2*3^2*4^2*5^2) is expressed as a decimal, how many distinct  [#permalink]

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Re: If x= 1/(2^2*3^2*4^2*5^2) is expressed as a decimal, how many distinct   [#permalink] 27 Sep 2019, 03:51
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