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26 Apr 2026, 04:33
Kudos
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
39% (03:07) correct
61%
(01:57)
wrong
based on 23
sessions
History
Date
Time
Result
Not Attempted Yet
What is the product of all positive odd integers less than 2500?
(A) \(\frac{2500!}{(1250!)^2}\)
(B) \(\frac{2500!}{2^{1250}}\)
(C) \(\frac{2499!}{2^{1250}}\)
(D) \(\frac{2500!}{2^{1250} \cdot 1250!}\)
(E) \(\frac{1250!}{2^{1250}}\)
(A) \(\frac{2500!}{(1250!)^2}\)
(B) \(\frac{2500!}{2^{1250}}\)
(C) \(\frac{2499!}{2^{1250}}\)
(D) \(\frac{2500!}{2^{1250} \cdot 1250!}\)
(E) \(\frac{1250!}{2^{1250}}\)
26 Apr 2026, 05:23
Kudos
Bookmarks
Answer is (D): 2500! / (2^1250 * 1250!)
This is a Number Properties question testing factorial manipulation. Took me a few tries to nail this type consistently.
The product of all positive odd integers less than 2500 is: 1 * 3 * 5 * 7 * ... * 2499.
Here's the trick. Notice that 2500! = (1 * 2 * 3 * 4 * ... * 2499 * 2500), which includes BOTH odd and even integers.
1. Split 2500! into its odd and even parts:
2500! = (1 * 3 * 5 * ... * 2499) * (2 * 4 * 6 * ... * 2500)
2. Factor out 2 from each even term:
(2 * 4 * 6 * ... * 2500) = 2^1250 * (1 * 2 * 3 * ... * 1250) = 2^1250 * 1250!
(There are 1250 even numbers from 2 to 2500, so we pull out 1250 factors of 2.)
3. So: 2500! = (product of odd integers 1 to 2499) * 2^1250 * 1250!
4. Solving for the product of odd integers:
Product = 2500! / (2^1250 * 1250!)
That's answer choice (D).
The trap is (A): 2500! / (1250!)^2 -- that's actually the binomial coefficient C(2500, 1250), which is a completely different thing. A lot of people see the 1250 denominator and jump to it.
Also watch (C): 2499! / 2^1250. This misses the 1250! factor entirely.
Key takeaway: when you need just the odd part of n!, factor the evens out explicitly and use the pattern above.
This is a Number Properties question testing factorial manipulation. Took me a few tries to nail this type consistently.
The product of all positive odd integers less than 2500 is: 1 * 3 * 5 * 7 * ... * 2499.
Here's the trick. Notice that 2500! = (1 * 2 * 3 * 4 * ... * 2499 * 2500), which includes BOTH odd and even integers.
1. Split 2500! into its odd and even parts:
2500! = (1 * 3 * 5 * ... * 2499) * (2 * 4 * 6 * ... * 2500)
2. Factor out 2 from each even term:
(2 * 4 * 6 * ... * 2500) = 2^1250 * (1 * 2 * 3 * ... * 1250) = 2^1250 * 1250!
(There are 1250 even numbers from 2 to 2500, so we pull out 1250 factors of 2.)
3. So: 2500! = (product of odd integers 1 to 2499) * 2^1250 * 1250!
4. Solving for the product of odd integers:
Product = 2500! / (2^1250 * 1250!)
That's answer choice (D).
The trap is (A): 2500! / (1250!)^2 -- that's actually the binomial coefficient C(2500, 1250), which is a completely different thing. A lot of people see the 1250 denominator and jump to it.
Also watch (C): 2499! / 2^1250. This misses the 1250! factor entirely.
Key takeaway: when you need just the odd part of n!, factor the evens out explicitly and use the pattern above.
