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# How many zeros will the decimal equivalent of

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How many zeros will the decimal equivalent of [#permalink]  30 Jul 2017, 08:27
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Question Stats:

36% (01:36) correct 63% (01:20) wrong based on 19 sessions

How many zeros will the decimal equivalent of $$\frac{1}{2^{11} * 5^7}+\frac{1}{2^7 *5^{11}}$$ have after the decimal point prior to the first non-zero digit?

(A) 6

(B) 7

(C) 8

(D) 11

(E) 18
[Reveal] Spoiler: OA

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Re: How many zeros will the decimal equivalent of [#permalink]  19 Sep 2017, 09:26
Any help here? I tried using the properties of exponential but I am completely stuck!
Director
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Re: How many zeros will the decimal equivalent of [#permalink]  22 Sep 2017, 20:32
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Carcass wrote:
How many zeros will the decimal equivalent of 1/{2^(11) * 5^(7)} + 1/{2^(7) *5^(11)} have after the decimal point prior to the first non-zero digit?

(A) 6

(B) 7

(C) 8

(D) 11

(E) 18

There is graphical representation error, even I read incorrectly i read as 1/2^1 * 15^7 but it is 1/{(2^11) * (5^7)} .Hope it helps

Now comming back to question to make the decimal terminating we have to bring the decimal in the form $$2^n * 5^n$$

From the equation $$\frac{1}{2^{11} * 5^7}+ \frac{1}{2^7 * 5^{11}}$$

We take$$\frac{1}{(2^7 * 5^7)} * [\frac{1}{(2^4)} + \frac{1}{(5^4)}]$$

= $$\frac{1}{10^7}$$ $$[\frac{1}{16}$$ + $$\frac{1}{625}]$$
= $$\frac{1}{10^7}$$ $$[\frac{641}{10^4}]$$
= $$\frac{641}{10^{11}}$$
= 8 decimal point ( since 641 >100)
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Director
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Re: How many zeros will the decimal equivalent of [#permalink]  25 Sep 2017, 02:51
pranab01 wrote:
Carcass wrote:
How many zeros will the decimal equivalent of 1/{2^(11) * 5^(7)} + 1/{2^(7) *5^(11)} have after the decimal point prior to the first non-zero digit?

(A) 6

(B) 7

(C) 8

(D) 11

(E) 18

There is graphical representation error, even I read incorrectly i read as 1/2^1 * 15^7 but it is 1/{(2^11) * (5^7)} .Hope it helps

Now comming back to question to make the decimal terminating we have to bring the decimal in the form $$2^n * 5^n$$

From the equation $$\frac{1}{2^{11} * 5^7}+ \frac{1}{2^7 * 5^{11}}$$

We take$$\frac{1}{(2^7 * 5^7)} * [\frac{1}{(2^4)} + \frac{1}{(5^4)}]$$

= $$\frac{1}{10^7}$$ $$[\frac{1}{16}$$ + $$\frac{1}{625}]$$
= $$\frac{1}{10^7}$$ $$[\frac{641}{10^4}]$$
= $$\frac{641}{10^{11}}$$
= 8 decimal point (since 641 >100)

Fine, the error in the formula completely drove me out way.

Just a question: I am not sure I have understood your final line "= 8 decimal point (since 641 >100)". I would have said that
$$\frac{641}{10^{11}}=164*10^{-11}$$, thus we have to move the point by 11 places behind and given that three places are numbers different from zero, there are 8 zeros before six, 11-3 = 8
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Re: How many zeros will the decimal equivalent of [#permalink]  25 Sep 2017, 04:23
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IlCreatore wrote:

Fine, the error in the formula completely drove me out way.

Just a question: I am not sure I have understood your final line "= 8 decimal point (since 641 >100)". I would have said that
$$\frac{641}{10^{11}}=164*10^{-11}$$, thus we have to move the point by 11 places behind and given that three places are numbers different from zero, there are 8 zeros before six, 11-3 = 8

This is because 641>100 and it is divisible by 100, so we need not account for that

we can leave the 3 digit because it will be greater than 100.

For example if we had $$6*10^{-11}$$ = 10 decimal point before nonzero , $$64*10 ^{-11}$$ = 9 decimal point before nonzero

You can try that on calculator
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Re: How many zeros will the decimal equivalent of   [#permalink] 25 Sep 2017, 04:23
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