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Bunuel
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If p=125×243×16/405, how many digits are in p? 15 Apr 2016, 01:28
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D
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25% (medium)

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77% (01:33) correct 23% (01:31) wrong based on 198 sessions

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If p=125×243×16/405, how many digits are in p?

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D
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If p=125×243×16/405, how many digits are in p? 15 Apr 2016, 03:46
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Bunuel
If p=125×243×16/405, how many digits are in p?

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Its D...4 digits. Value will be in 1xxx.
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If p=125×243×16/405, how many digits are in p? 15 Apr 2016, 11:59
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Bunuel
If p=125×243×16/405, how many digits are in p?

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\(p\) = \(125×243×\frac{16}{405}\)

\(p\) = \(25×243×\frac{16}{81}\)

\(p\) = \(25×3×16\)

\(p\) = \(25×48\)

\(p\) = \(1200\) { Shortcut calculation is 4800/4 , since 25 = 100/4; add 2 zeroes to 48 then divide by 4}

Thus answer must be D. 4 :lol:
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If p=125×243×16/405, how many digits are in p? 15 Apr 2016, 19:52
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Bunuel
If p=125×243×16/405, how many digits are in p?

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We can or rather we SHOULD do some approximation..
the fraction 16/405 in a way is given to approximate, otherwise it could be 19/405 or so..
so lets take 405 as 400..

\(p=125×243×\frac{16}{400} = \frac{25 * 5 *243 *4*4}{4*4*25}\)..
=> 5*243 = slightly >1000 so 4 digits
D
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If p=125×243×16/405, how many digits are in p? 15 Apr 2016, 23:09
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Bunuel
If p=125×243×16/405, how many digits are in p?

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Approximation is your best bet here. Note that 405 is a bit less than twice of 243. It's between 1.5 and 2 times of 243.

p = 125 * 16/2 = 125 * 8 = 1000

So p is at least 1000. It is a bit more than 1000. So it must have 4 digits.

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If p=125×243×16/405, how many digits are in p? 16 Apr 2016, 00:25
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p = 125 * 243 * 16 / 405
p= 5^3 * 3 * 9^2 * 4^2 / (5 * 9^2)
p= 5^2 * 3 * 4^2
p = 20^2 * 3 = 1200

answer D

I've been wrong few times while working with powers of 10 or with approximations... in this example it's quite easy but sometimes the numbers are much complexe, so I don't feel comfortable if I d'ont go thru the numbers.. And considering that these type of questions are supposed to be the "easy time saving question" the test day, the risk of inattention errors increased... :?
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If p=125×243×16/405, how many digits are in p? 04 Feb 2020, 06:06
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P = 125*243*(16/405) = (5^3 *3^5*2^4)/(5*3^4) = 5^2*3*2^4=10^2*12

For 10, Digits 2+1 =3
For 12, Digits,1+0 =1
Thus total digits would be (3+1) = 4
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If p=125×243×16/405, how many digits are in p? 19 Jan 2025, 16:25
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