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

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Director
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Joined: 09 Mar 2016
Posts: 608
Re: If d=1/(2^3*5^7) is expressed as a terminating decimal, how [#permalink]

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New post 31 Dec 2017, 06:12
[/quote]

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?[/quote]

Hello Bunuel, i have the same question: based on which rule are you multiplying (2^3*5^7) by 2^4 but still ignoring 5^7 ? please help me to understand your smart solution :-) and to answer your question: if multiply (2^3*5^7) by 2^4 I get 2^7 *10^11 Thanks!
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If d=1/(2^3*5^7) is expressed as a terminating decimal, how [#permalink]

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New post 31 Dec 2017, 08:24
1
dave13 wrote:


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?[/quote]

Hello Bunuel, i have the same question: based on which rule are you multiplying (2^3*5^7) by 2^4 but still ignoring 5^7 ? please help me to understand your smart solution :-) and to answer your question: if multiply (2^3*5^7) by 2^4 I get 2^7 *10^11 Thanks![/quote]

Hi dave13

Properties of exponents say that if base is equal then on multiplying you add the powers and on dividing you subtract the powers

i.e. \(a^b*c^d\) if multiplied by \(a^x\), then it will become \(a^{(b+x)}*c^d\), here there will be no impact on \(c^d\) which has a different base.

In this problem as \(5^7\) has a power of \(7\) so we need to make \(2^3\) as \(2^7\), hence we multiply the numerator & denominator by \(2^4\)

so \(2^3*5^7*2^4=2^{(3+4)}*5^7=2^7*5^7=(2*5)^7=10^7\)
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Joined: 11 Sep 2017
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Re: If d=1/(2^3*5^7) is expressed as a terminating decimal, how [#permalink]

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New post 31 Dec 2017, 14:08
ilovefrankee wrote:
(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.


Hi Ben,

I'd like to highlight a few things-

1. The two non zero digits are ONE and SIX and not 2 & 5.

2. In your manipulation of the expression, notice that you have missed out a TWO in the 3rd step (1 / (2*(2^2))*(5^7) =
1 / (4x10^7) and where did you get that TEN (1 / (2*(2^2))*(5^7) =
1 / (4x10^7)) from?

Correct manipulation will be: 1 / (2^3*5^7) = 1 / (2^3*5^7) * 2^4/ 2^4 = 2^4/ 10^7 = 16/ 10^7 = 0.0000016. Hence, the two non zero digits are 1 and 6.

Hope this helps.

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

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New post 02 Jan 2018, 03:58
niks18 wrote:
dave13 wrote:


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?


Hello Bunuel, i have the same question: based on which rule are you multiplying (2^3*5^7) by 2^4 but still ignoring 5^7 ? please help me to understand your smart solution :-) and to answer your question: if multiply (2^3*5^7) by 2^4 I get 2^7 *10^11 Thanks![/quote]

Hi dave13

Properties of exponents say that if base is equal then on multiplying you add the powers and on dividing you subtract the powers

i.e. \(a^b*c^d\) if multiplied by \(a^x\), then it will become \(a^{(b+x)}*c^d\), here there will be no impact on \(c^d\) which has a different base.

In this problem as \(5^7\) has a power of \(7\) so we need to make \(2^3\) as \(2^7\), hence we multiply the numerator & denominator by \(2^4\)

so \(2^3*5^7*2^4=2^{(3+4)}*5^7=2^7*5^7=(2*5)^7=10^7\)[/quote]

Thanks a lot niks18 for a detailed explanation. Highly appreciated!
Re: If d=1/(2^3*5^7) is expressed as a terminating decimal, how   [#permalink] 02 Jan 2018, 03:58

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