If a = 1/2 * 3/4 * ... * 99/100, is a

Question

Try this one guys:

If a = 1/2 * 3/4 * ... * 99/100, is a 1/10

FN · Answer

B it is...

it seems the series in a keeps getting close to 1, but never quite reaches there...0.99 is as close it gets to 1...so we know the result of the multiplication is going to be way less than 1,

Statement 1) say b is between 0 and 1, which is what I expect series a to be...so therefor insufficient to tell if B > A.

Statement2) is clearly greater than 1/10 and we know that the series of A is less than 1/10. B is sufficient.

MA · Answer

From i, we know the value of a but do not know the value of b. therefore, we can say a and b, both, are smaller than 1 and greater than 0 but we do not know which one is greater and which one is smaller. 

from ii, the value of b is more than 1/10, and the value of a 


a= 1/2*3/4=0.38 
a= 1/2*3/4*5/6=0.31 
a= 1/2*3/4*5/6*7/8=0.27 
a= 1/2*3/4*5/6*7/8*9/10=0.25 
a= 1/2*3/4*5/6*7/8*9/10*11/12=0.21 

the value of a is in decreasing order. from the above simulation, we can generalize that a's value will decease as we add more terms. so, a will be less than 1/10. therfore it is safe to say b>a.

HongHu · Answer

This question is quite dangerous, in my mind. I used Excel to look at the series. It only gets to be smaller then 0.1 after the 63 term. I guess all we could do is to guess. If the series is ended at 49/50 instead of 99/100, then we'd be wrong. Anybody has any good way to determine that it is indeed less than 0.1 rather than by guessing?

speedo · Answer

Not the 1st time me hearing these Excel goof ups!! :lol:

Just try and find out the value 3^38 using your Excel, you'll understand what I mean. :twisted:

& probably stop using Excel from now on!! Use your own Quant skills, will give more accurate results. I think fresinha12 has pretty much summed it up. :wink:

qhoc0010 · Answer

Very good!!

OK, I guess you guys cannot wait for the OA since this one is hard. The OA is (B)

Let solve this one.

(1) With common sense, we probably can conclude right the way that this one is insufficient.
Obviously, 0 < a < 1
Also 0 < b < 1
a is a constant and b is variable so (1) will give you nothing.
b could be 0.00000000000000001 or 0.999999999999999999

(2) OK, the second one gives you the number 1/10
You may want to find out what this number is for.
If (2) is sufficient, a = 1/10 or a < 1/10

Look at the pattern in a: k/(k + 1), where k is an integer greater than 0
Let
c = 1/2 * 3/4 * ... * 97/100 (1)
-> a = c * 99/100

Notice that:
k/(k + 1) < (k + 1)/(k + 2)

Quick proof here:
k/(k + 1) < (k + 1)/(k + 2)
-> k * (k + 2) < (k + 1) * (k + 1)
-> k^2 + 2*k < k^2 + 2*k + 1
-> 0 < 1 (TRUE)

According to the inequality above:
1/2 < 2/3
1/2 * 3/4 < 2/3 * 4/5
...

We can say:
c < 2/3 * 4/5 * ... * 98/99 (2)

Multiply (1) and (2)

c^2 < 1/2 * 2/3 * ... * 97/98 * 98/99
c^2 < 1/99

-> c < 1/sqrt(99)

We have:
a = c * 99/100 <1/sqrt(99) * 99/100 = sqrt(99)/100 < 10/100 = 1/10
-> a < 1/10

WOW, done. This mean if b > 1/10, then a is always less than b since a < 1/10

By the way, I made up this question!!! :lol:

nocilis · Answer

qhoc0010's answer makes sense. It is quite complicated and not possible to do in 2 minutes.
I agree with HongHu that other solutions have a leap-of-faith arguments rather than arguments with a solid justification. 

In general, I am troubled by the fact that quite a few DS problems involving number theory involve such a leap-of-faith solution because we need to solve the problem in two minutes! (try three numbers, if it works, then it works, if one of them do not work, solution is not good - this is clearly a leap-of-faith)

If a = 1/2 * 3/4 * ... * 99/100, is a < b ?

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