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

A. 1
B. 2
C. 3
D. 4
E. 5

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

A. 1
B. 2
C. 3
D. 4
E. 5

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.

Answer (D)
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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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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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