What is the product of all positive odd integers less than 10000?

Half of the positive integers up to 10,000 are even, so there are 5000 positive even integers up to 10,000. If we list them, in descending order, we have

10000, 9998, 9996, ..., 6, 4, 2

From each, we can factor out a 2, so we can rewrite the numbers above in this way:

(2)(5000), (2)(4999), (2)(4998), ..., (2)(3), (2)(2), (2)(1)

Now if we multiply these 5000 things together, we have all 5000 of the 2's that we just factored out, giving us 2^5000, along with the product of the other numbers, 5000*4999*4998*...*3*2*1 = 5000!. So the product of the positive even numbers up to 10,000 is equal to 2^5000 * 5000!. The product of all of the positive integers up to 10,000 is 10,000!, so if we cancel the even numbers from 10,000!, by dividing by 2^5000*5000!, that will leave us with the product of the odd numbers up to 10,000.

[quote="nick1816"]10000!= 1*2*3*4*.....*9999*10000

10000!= (1*3*5*....*9997*9999) * (2*4*6.......*9998*10000)

10000!= (1*3*5*....*9997*9999) * \(2^{5000}\)(1*2*3.......*4999*5000)

10000!= (1*3*5*....*9997*9999) * \(2^{5000}\)*(5000!)

(1*3*5*....*9997*9999)= \(\frac{10000!}{ 2^{5000}*(5000!)}\)



[quote="Bunuel"]What is the product of all positive odd integers less than 10000?

I understand the math you did but how did you realize to do this???

tried with smaller nos
let 10000 ---> 8
then odd nos are 1 3 5 and 7 , which when multiplied give 105.

D results in the same.

Asked: What is the product of all positive odd integers less than 10000?

The product of all positive odd integers less than 10000 \(= 1*3*5*7*....*9999 = \frac{1*2*3*4*.....9999}{2*4*6*...9998} = \frac{9999!}{2^{4999}*4999!} = \frac{10000!}{2^{5000}*5000!}\\
\\
IMO D\\
\)

followed this approach, not sure if it will be applicable to all such ques:

- Take a sample of first 10 integers,

- Product of odd integers from first 10: 9*7*5*3*1 = 63*15

Now, Go through options and make similar scenario:

Option A: \(\frac{10000!}{(5000!)^2}\) --> Similar to it, \(\frac{10!}{(5!)^2} = 63*4 .... No\)

Same way...Option B...no

Option C. ..no

Option D: \(\frac{10000!}{2^(5000) .5000!}\) --> Similar to it, \(\frac{10!}{(2)^5.5!} =63*15\)

IMO Option D!

I do understand the math behind it, but I don't understand why the answers are in fractions if we are talking about the product of a number?

Can the product of the 2 numbers equal to \(\frac{6}{2} \)? Obviously yes, because reduced form of the fraction 6/2 is an integer (3).

Same thing is going on in this question.

\(\frac{10000!}{2^{5000}. 5000!}\)

= \(\frac{5001*5002* 5003*......*9999*10000}{2^{5000}}\)

Now let's check the maximum power of 2, that can completely divide the numerator. If it's greater than or equal to 5000, the reduced form of the fraction is an integer.

The maximum power of 2, that can completely divide the numerator

= ([10000/2 ]+ [10000/4]+[10000/8]...........) - ([5000/2]+[5000/4]+...........)

= ([10000/2 ]+ [10000/4]+[10000/8]...........) - ([10000/4]+[10000/8]+...........)

> [10000/2 ] = 5000

Hence, the reduced form of the above fraction is an integer.

Ok! But in the third line, there's something I did not understand.

You said that 10000! =(1*3*5*....*9997*9999), which is all odd numbers times \(2^{5000}\)(1*2*3.......*4999*5000)
Why both \(2^{5000}\) and * (1*2*3.......*4999*5000)) ?

Should not it be the first term (odd numbers) times 2(2*4*6*8....), which is \(2^{5000}\) .

Thank you!

I'm not getting the passages, could you explain every step in detail PLEASE'?

can someone explain to me why we can write the denominator as 2 to 5000 (how do we know there are exactly 5000 2's? I dont' see it) MULTIPLIED BY 5000!
thanks

Bunuel, do you know the source of this question? I published a nearly identical question roughly ten years ago, in my Problem Set #3, though using much smaller numbers and different wording.

let us first take a simple case and then expand the generalisation

for the first 10 terms it's equal to 10! / (2^5*5!)

When we look at the options and check we get D
THerefore IMO D

The question is from American Mathematics Competitions 2012: https://artofproblemsolving.com/wiki/in ... /Problem_5

YOUR EXPLANATION WAS PERFECTLY CLEAR. THANK YOU SO MUCH, REALLY. (:

The product of all positive odd integers that are ess than 10000.

The numbers include 1,3, 5,........................9999.

The product of ( 1*3*5*..........................9999).

Multiplying the even numbers starting from 2,.....................10000 and dividing with the same.

We have : \(\frac{\left(1\cdot2\cdot3\cdot....................10000\right)}{\left(2\cdot4\cdot................10000\right)}\)

The numerator can be rewritten as \(10000!\)

For the denominator, we can take 2 common from each term and the denominator becomes :

\(\left(2^{5000}\right)\cdot\left(5000!\right)\\
\)
The value becomes equal to :

\(\frac{\left(10000!\right)}{\left(2^{5000}\right)\cdot\left(5000!\right)}\)

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