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27 Jan 2017, 16:43
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Bunuel
"Now, does \(p\) divided by any of these \(d's\) have fewer than 5 decimal places? Yes, as \(\frac{p}{d}*10,000=integer\) for any such \(d\) (10,000 is divisible by all these numbers: 1, 2, 4, 5, 8, 10, 16, 20, 25)."
p/d*100,000=integer as well, for any of the d's. So how can you tell it does not have 5 decimal places? I don't understand the rationale.
CholericQuill
Joined: 06 Jan 2016
Last visit: 18 Apr 2017
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Location: Australia
Concentration: Strategy, Statistics
GMAT 1: 720 Q48 V41
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GMAT 1: 720 Q48 V41
Posts: 4
27 Jan 2017, 18:42
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plalud
A number that has less than 5 decimal places can be represented as \(\frac{k}{10,000}\), where k is an integer.
Essentially, \(\frac{p}{d}\)=\(\frac{k}{10,000}\), if \(\frac{p}{d}\) has less than 5 decimal places.
This is the rationale behind multiplying \(\frac{p}{d}\) with 10,000 to check if the product is an integer. Since, 10,000 is divisible by all possible denominators, \(\frac{p}{d}*10,000\) is an integer.
Of course, \(\frac{p}{d}*100,000\) is also an integer as you pointed out, but as we have already established \(\frac{p}{d}*10,000\) is an integer, \(\frac{p}{d}*100,000\) will be a multiple of 10 (The 5th decimal place is 0)
Hope that helps!
24 May 2021, 19:23
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It took me a bit to understand this question when I first saw it, along with the accompanying Veritas Blog on Terminating and Recurring Decimals.
One of the things I started to see throughout all of this...studying....is that understanding numbers and why things are they way they are really helps to understand the more difficult problems.
When we did long division in grade school, we never really thought about why we add additional 0s after the decimal point when we complete long division to find that Quotient.
First, understand that a fraction in which the reduced form has only the Prime Factors of (2) and (5) will produce a terminating decimal.
Why this is so?
(1 way to look at)
As someone stated above, if we can re-write the fraction in a form such that the DEN = power of 10, then the rule for (-)negative integer powers:
P / (10)^4 =same as= (P) * (10)^-4
Which means regardless of the Prime integer P, we will move the decimal point 4 places to the left.
101 ———> becomes 0 . 0101
So for each day that has only prime bases or (2) and/or (5) we can multiply the NUM and DEN by the missing 5s or 2s that we need to make the DEN a Power or 10
For example, let P = 101 = sum up to day 20
And we have Day = d = 20 = (2)^2 * (5)^1
The average of deposits up through day 20 would be:
101 / (2)^2 * (5)^1
We can multiply the top and bottom of this fraction by (5)^1 to complete the power of 10 in the DEN ——-> * (5^1) / (5^1) = equivalent to multiplying by * 1
(101) * (5^1)
_________
(2)^2 * (5)^1 * (5^1)
Which ends up being
505
_______
(10)^2
= (505) * (10)^-2
= 5.05
Only 2 decimal places
(2nd way to understand the concept and problem)
I’m using d = 4 to illustrate the next way to understand it but you can use any other of the 9 days that work in Bunuel’s solution.
On day 4, the average up to that day will be:
P / 4 = (P/1) * (1/4)
P/1 will produce an integer result obviously. How many decimal points there will be in the result depends on how many decimal points are in (1/4)
Then you can think about what we are actually doing when we perform long division.
For a fraction such as (1/4):
We would take the NUM of 1 and put it under the division line. Then we put the Divisor of 4 outside the lines and “add 0s” after the decimal point that follows 1, until we reached our terminating decimal.
When we are “adding the zeros after the decimal point” in long division, we can think of this as multiplying the (1/4)’s NUM and DEN by Powers of 10 that will accommodate the Divisor so that 4 will divide evenly into the power of 10
The least power of 10 is the minimum number of zeros we would have to add after the NUM of 1’s decimal point until we get a terminating decimal
For (2)^2——-> we will need to add two zeros after the decimal (a 10^2 for the 2^2 to divide evenly into)
this will work for any NUM that is a NON-multiple of 4 divided by 4.
For 1 as the NUM———> we make the 1 under the division line = 1.00
And the decimal will terminate within those 2 places.
Further, for divisor 4, the integer remainders can be: 1 , 2 , 3
For P/4 ——-> the quotient decimal result will take the form of any of the following:
XX .25
XX .50
XX .75
Because the: (Integer Rem) = (Divisor) * ( Decimal Portion of Quotient)
(4) (.25) = 1
(4) (.50) = 2
(4) (.75) = 3
In summary, for the DEN = d = (2)^2
For a non-multiple of 4 in the NUM we will only need “two more zeros” added to the decimal point under the long division symbol.
Or
You can say:
(P / (2)^2) * (10)*2 = Integer
You can work through other examples to see this in action by making P = any prime greater than 100 and then playing with the various denominators that work in Bunuel’s solution.
Seeing it work will provide a better understanding.
Posted from my mobile device
One of the things I started to see throughout all of this...studying....is that understanding numbers and why things are they way they are really helps to understand the more difficult problems.
When we did long division in grade school, we never really thought about why we add additional 0s after the decimal point when we complete long division to find that Quotient.
First, understand that a fraction in which the reduced form has only the Prime Factors of (2) and (5) will produce a terminating decimal.
Why this is so?
(1 way to look at)
As someone stated above, if we can re-write the fraction in a form such that the DEN = power of 10, then the rule for (-)negative integer powers:
P / (10)^4 =same as= (P) * (10)^-4
Which means regardless of the Prime integer P, we will move the decimal point 4 places to the left.
101 ———> becomes 0 . 0101
So for each day that has only prime bases or (2) and/or (5) we can multiply the NUM and DEN by the missing 5s or 2s that we need to make the DEN a Power or 10
For example, let P = 101 = sum up to day 20
And we have Day = d = 20 = (2)^2 * (5)^1
The average of deposits up through day 20 would be:
101 / (2)^2 * (5)^1
We can multiply the top and bottom of this fraction by (5)^1 to complete the power of 10 in the DEN ——-> * (5^1) / (5^1) = equivalent to multiplying by * 1
(101) * (5^1)
_________
(2)^2 * (5)^1 * (5^1)
Which ends up being
505
_______
(10)^2
= (505) * (10)^-2
= 5.05
Only 2 decimal places
(2nd way to understand the concept and problem)
I’m using d = 4 to illustrate the next way to understand it but you can use any other of the 9 days that work in Bunuel’s solution.
On day 4, the average up to that day will be:
P / 4 = (P/1) * (1/4)
P/1 will produce an integer result obviously. How many decimal points there will be in the result depends on how many decimal points are in (1/4)
Then you can think about what we are actually doing when we perform long division.
For a fraction such as (1/4):
We would take the NUM of 1 and put it under the division line. Then we put the Divisor of 4 outside the lines and “add 0s” after the decimal point that follows 1, until we reached our terminating decimal.
When we are “adding the zeros after the decimal point” in long division, we can think of this as multiplying the (1/4)’s NUM and DEN by Powers of 10 that will accommodate the Divisor so that 4 will divide evenly into the power of 10
The least power of 10 is the minimum number of zeros we would have to add after the NUM of 1’s decimal point until we get a terminating decimal
For (2)^2——-> we will need to add two zeros after the decimal (a 10^2 for the 2^2 to divide evenly into)
this will work for any NUM that is a NON-multiple of 4 divided by 4.
For 1 as the NUM———> we make the 1 under the division line = 1.00
And the decimal will terminate within those 2 places.
Further, for divisor 4, the integer remainders can be: 1 , 2 , 3
For P/4 ——-> the quotient decimal result will take the form of any of the following:
XX .25
XX .50
XX .75
Because the: (Integer Rem) = (Divisor) * ( Decimal Portion of Quotient)
(4) (.25) = 1
(4) (.50) = 2
(4) (.75) = 3
In summary, for the DEN = d = (2)^2
For a non-multiple of 4 in the NUM we will only need “two more zeros” added to the decimal point under the long division symbol.
Or
You can say:
(P / (2)^2) * (10)*2 = Integer
You can work through other examples to see this in action by making P = any prime greater than 100 and then playing with the various denominators that work in Bunuel’s solution.
Seeing it work will provide a better understanding.
Posted from my mobile device
