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Re: How many zeros will the decimal equivalent of [#permalink]
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]
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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 :P
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How many zeros will the decimal equivalent of [#permalink]
pranab223 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)


How do you get from [(1/16)+(1/624)] to [641/(10^4)]
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