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If d=1/(2^3*5^7) is expressed as a terminating decimal, how

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If d=1/(2^3*5^7) is expressed as a terminating decimal, how [#permalink]

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20 Dec 2012, 06:11
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If d=1/(2^3*5^7) is expressed as a terminating decimal, how many nonzero digits will d have?

(A) One
(B) Two
(C) Three
(D) Seven
(E) Ten
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Re: If d=1/(2^3*5^7) is expressed as a terminating decimal, how [#permalink]

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20 Dec 2012, 06:12
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If d=1/(2^3*5^7) is expressed as a terminating decimal, how many nonzero digits will d have?

(A) One
(B) Two
(C) Three
(D) Seven
(E) Ten

Given: $$d=\frac{1}{2^3*5^7}$$.

Multiply by $$\frac{2^4}{2^4}$$ --> $$d=\frac{2^4}{(2^3*5^7)*2^4}=\frac{2^4}{2^7*5^7}=\frac{2^4}{10^7}=\frac{16}{10^7}=0.0000016$$. Hence $$d$$ will have two non-zero digits, 16, when expressed as a decimal.

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Re: If d=1/(2^3*5^7) is expressed as a terminating decimal, how [#permalink]

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24 Oct 2013, 03:15
1/2^3*5^7 = 2^-3*5^-7 =.002 * .0000007. So there are 2 non zero digits!!

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Re: If d=1/(2^3*5^7) is expressed as a terminating decimal, how [#permalink]

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24 Oct 2013, 03:28
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Puneethrao wrote:
1/2^3*5^7 = 2^-3*5^-7 =.002 * .0000007. So there are 2 non zero digits!!

Unfortunately this is not correct:

$$2^{-3}=\frac{1}{8}=0.125$$ not 0.002, which is 2/10^3 and $$5^{-7}=\frac{1}{78,125}=0.0000128$$ not 0.0000007, which is 7/10^7.

Hope it helps.
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Re: If d=1/(2^3*5^7) is expressed as a terminating decimal, how [#permalink]

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24 Oct 2013, 03:53
Bunuel wrote:
Puneethrao wrote:
1/2^3*5^7 = 2^-3*5^-7 =.002 * .0000007. So there are 2 non zero digits!!

Unfortunately this is not correct:

$$2^{-3}=\frac{1}{8}=0.125$$ not 0.002, which is 2/10^3 and $$5^{-7}=\frac{1}{78,125}=0.0000128$$ not 0.0000007, which is 7/10^7.

Hope it helps.

Thanks a lot!! I don't know what i was thinking , such a stupid mistake!! Thanks once again!

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Re: If d=1/(2^3*5^7) is expressed as a terminating decimal, how [#permalink]

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29 Dec 2013, 12:57
Bunuel wrote:
If d=1/(2^3*5^7) is expressed as a terminating decimal, how many nonzero digits will d have?

(A) One
(B) Two
(C) Three
(D) Seven
(E) Ten

Given: $$d=\frac{1}{2^3*5^7}$$.

Multiply by $$\frac{2^4}{2^4}$$ --> $$d=\frac{2^4}{(2^3*5^7)*2^4}=\frac{2^4}{2^7*5^7}=\frac{2^4}{10^7}=\frac{16}{10^7}=0.0000016$$. Hence $$d$$ will have two non-zero digits, 16, when expressed as a decimal.

I have seen couple of more problem like this. One thing is still not clear to me. When you multiply whole denominator by 2^4 why is 5^7 getting ignored? Shouldn't 2^4 multiply both 2^3 as well as 5^7?

Thanks

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Re: If d=1/(2^3*5^7) is expressed as a terminating decimal, how [#permalink]

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29 Dec 2013, 13:00
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theGame001 wrote:
Bunuel wrote:
If d=1/(2^3*5^7) is expressed as a terminating decimal, how many nonzero digits will d have?

(A) One
(B) Two
(C) Three
(D) Seven
(E) Ten

Given: $$d=\frac{1}{2^3*5^7}$$.

Multiply by $$\frac{2^4}{2^4}$$ --> $$d=\frac{2^4}{(2^3*5^7)*2^4}=\frac{2^4}{2^7*5^7}=\frac{2^4}{10^7}=\frac{16}{10^7}=0.0000016$$. Hence $$d$$ will have two non-zero digits, 16, when expressed as a decimal.

I have seen couple of more problem like this. One thing is still not clear to me. When you multiply whole denominator by 2^4 why is 5^7 getting ignored? Shouldn't 2^4 multiply both 2^3 as well as 5^7?

Thanks

Frankly, the red part does not make any sense...

The denominator is $$2^7*5^7$$. Multiply it by $$2^4$$. What do you get?
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Re: If d=1/(2^3*5^7) is expressed as a terminating decimal, how [#permalink]

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11 Mar 2014, 16:54
Bunuel wrote:
If d=1/(2^3*5^7) is expressed as a terminating decimal, how many nonzero digits will d have?

(A) One
(B) Two
(C) Three
(D) Seven
(E) Ten

Given: $$d=\frac{1}{2^3*5^7}$$.

Multiply by $$\frac{2^4}{2^4}$$ --> $$d=\frac{2^4}{(2^3*5^7)*2^4}=\frac{2^4}{2^7*5^7}=\frac{2^4}{10^7}=\frac{16}{10^7}=0.0000016$$. Hence $$d$$ will have two non-zero digits, 16, when expressed as a decimal.

What is it that you saw that indicated you should multiply by 2^4. Just looking at the problem that never occurred to me and I'd like to understand why it did to you.

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Re: If d=1/(2^3*5^7) is expressed as a terminating decimal, how [#permalink]

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11 Mar 2014, 23:36
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WinterIsComing wrote:
Bunuel wrote:
If d=1/(2^3*5^7) is expressed as a terminating decimal, how many nonzero digits will d have?

(A) One
(B) Two
(C) Three
(D) Seven
(E) Ten

Given: $$d=\frac{1}{2^3*5^7}$$.

Multiply by $$\frac{2^4}{2^4}$$ --> $$d=\frac{2^4}{(2^3*5^7)*2^4}=\frac{2^4}{2^7*5^7}=\frac{2^4}{10^7}=\frac{16}{10^7}=0.0000016$$. Hence $$d$$ will have two non-zero digits, 16, when expressed as a decimal.

What is it that you saw that indicated you should multiply by 2^4. Just looking at the problem that never occurred to me and I'd like to understand why it did to you.

We need to multiply by 2^6/2^6 in order to convert the denominator to the base of 10 and then to convert the fraction into the decimal form: 0.xxxx.

Similar questions to practice:
if-t-1-2-9-5-3-is-expressed-as-a-terminating-decimal-ho-129447.html
if-d-1-2-3-5-7-is-expressed-as-a-terminating-decimal-128457.html

Hope this helps.
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Re: If d=1/(2^3*5^7) is expressed as a terminating decimal, how [#permalink]

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27 May 2014, 03:29
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Another approach:

$$\frac{1}{(2^3*5^7)}$$ =$$\frac{1}{(2^3*5^3*5^4)}$$ by splitting denominator.

= $$\frac{1}{(10^3*5^4)}$$ = $$\frac{10^{-3}}{5^4}$$

Representing numerator as$$\frac{(10^4*10^{-7})}{5^4}$$ = $$2^4*10^{-7}$$ = $$16*10^{-7}$$

=.0000016 , Hence 2 digits.

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Re: If d=1/(2^3*5^7) is expressed as a terminating decimal, how [#permalink]

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05 Jan 2016, 07:03
2^7 * 1/(10^7) * 2^-3 = 2^4 * 1/(10^7) = 16/10000000 = .000000016

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Re: If d=1/(2^3*5^7) is expressed as a terminating decimal, how [#permalink]

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23 Jun 2016, 09:38
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If d=1/(2^3*5^7) is expressed as a terminating decimal, how many nonzero digits will d have?

(A) One
(B) Two
(C) Three
(D) Seven
(E) Ten

Since actually dividing 1/(2^3*5^7) would be time consuming, we want to manipulate d so that we are working with a cleaner denominator. The easiest way to do that is to multiply d by a value that will produce a perfect power of 10 in the denominator. This means that the number of 2s in the denominator will equal the number of 5s in the denominator.

Thus, we can multiply 1/(2^3*5^7) by 2^4/2^4. This gives us:

2^4/(2^7*5^7)

2^4/10^7

16/10^7

16/10,000,000

We can stop here because we know that the 10,000,000 in the denominator means to move the decimal place after the 16 seven places to the left. The final value of d will be 0.0000016. Note that the division of 16 by 10,000,000 did not produce any additional non-zero digits. Thus d has 2 non-zero digits.

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Re: If d=1/(2^3*5^7) is expressed as a terminating decimal, how [#permalink]

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23 Jun 2016, 11:35
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If d=1/(2^3*5^7) is expressed as a terminating decimal, how many nonzero digits will d have?

(A) One
(B) Two
(C) Three
(D) Seven
(E) Ten

$$d$$ = $$\frac{1}{(2^3*5^7)}$$

=>$$d$$ = $$\frac{1}{(2^3*5^3*5^4)}$$

=>$$d$$ = $$\frac{1}{(10^3*5^4)}$$

$$\frac{1}{5}$$ = $$0.20$$

$$\frac{1}{25}$$ = $$\frac{0.20}{5}$$ => $$0.04$$

$$\frac{1}{125}$$ = $$\frac{0.04}{5}$$ => $$0.008$$

$$\frac{1}{625}$$ = $$\frac{0.008}{5}$$ => $$0.0016$$

Hence there will be 2 non zero digits...

Feel free to revert in case of any doubt ( I have used some shortcuts , would love to explain if needed )

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Re: If d=1/(2^3*5^7) is expressed as a terminating decimal, how [#permalink]

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12 Sep 2016, 08:11
I solved the question in the following method, not sure whether it is correct:

1/2^3 x 5^7 = 1/2^3 x 5^3 [Equating the power of 2 & 5 to get the number of zeros], left with 1/5^4 = 1/625 = 0.00105. Only 1 & 5 are the non-zero digits.

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If d=1/(2^3*5^7) is expressed as a terminating decimal, how [#permalink]

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12 Sep 2016, 08:23
I did it this way.

d = $$\frac{1}{8*25*25*25*5}$$ = $$\frac{4*4*4*2}{8*100*100*100*10}$$ = 16 * 10 ^ -6 ==> 2 non zero digits.
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If d=1/(2^3*5^7) is expressed as a terminating decimal, how [#permalink]

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25 Jan 2017, 13:45
(First ever post!)

I realise I'm a little late submitting my answer here, but my answer was 2 non-zero digits: 2 & 5.

My answer is based on the following:

1 / (2^3*5^7) = 1 / (2*(2^2))*(5^7) =
1 / (4x10^7) =
25 x 10^8 .

I'm guessing my mistake was in factoring the denominator, specifically factoring of 2^3 as 2x2^2?

Any input greatly appreciated,

Ben

EDIT:

Not to worry, I've gone over some other exponent materials and came up with the correct solution.

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If d=1/(2^3*5^7) is expressed as a terminating decimal, how [#permalink]

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01 Apr 2017, 06:35
d=1/(2^3*5^7)

2*5=10 (Happy to know in such questions!!)
Write 2^3*5^3 together. (This will be equal to 2*2*2*5*5*5= 10^3)

d= 1/(10^3)(5^4)
d= [1/(10^3)] *(0.2)^4
d=[1/(10^3)] *(1.6)

Thus the non-zero digits in the final answer would be 2 i.e. 1 and 6.
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Re: If d=1/(2^3*5^7) is expressed as a terminating decimal, how [#permalink]

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29 Apr 2017, 11:47
1/[(2^3)(5^7)] = 1/(10^3)(5^4)
1/(10^3)=0.001
0.001/(5^4)=0.0000016

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27 May 2017, 07:17
Split the two fraction 1/2^7 * 1/5^7
1/5^7 = (1/5)^7 = 0.2^7 = (2*10^-1)^7 = 2^7 * 10^-7
Multiply by 1/2^3 now you get:
2^7/2^3 * 10^-7 = 2^4 * 10^-7 = 16*10^-7

The two non zero digits are then 1 and 6.

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Re: If d=1/(2^3*5^7) is expressed as a terminating decimal, how [#permalink]

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27 May 2017, 08:15
narendran1990 wrote:
I solved the question in the following method, not sure whether it is correct:

1/2^3 x 5^7 = 1/2^3 x 5^3 [Equating the power of 2 & 5 to get the number of zeros], left with 1/5^4 = 1/625 = 0.00105. Only 1 & 5 are the non-zero digits.

Check the highlighted part

$$\frac{1}{625} = 0.0016$$

There will be 2 non zero digits...

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Re: If d=1/(2^3*5^7) is expressed as a terminating decimal, how   [#permalink] 27 May 2017, 08:15
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