Hi there,
Could you give me your ELEGANT, QUICK, and EFFICIENT method to resolve (with these concepts and with finding the solution).
Which of the following is an integer?
1. 12!/6!
2. 12!/8!
3. 12!/(7!5!)
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Nermal
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The first two terms definitely each yield an integer.
12!/6! ---> 6! is part of 12!, so in the numerator you acutally only have the integers from 12 to 7 inclusive.
If you multiply any two integers you will always get an integer as the result.
Same reasoning for 12!/8!.
12!/(7!5!): a little trickier.
Treat it like 12!/7! first. Then we are left with 12*11*10*9*8/5!
Look for common prime factors of the numerator and the denominator.
We can rewrite 5! as 5*4*3*2*1=5*2*2*3*2=2*2*2*3*5
Are there three 2s, one 3 and one 5 in the numerator?
Yes, there are.
Therefore all of those expressions will yield an integer.
I,II and III
12!/6! ---> 6! is part of 12!, so in the numerator you acutally only have the integers from 12 to 7 inclusive.
If you multiply any two integers you will always get an integer as the result.
Same reasoning for 12!/8!.
12!/(7!5!): a little trickier.
Treat it like 12!/7! first. Then we are left with 12*11*10*9*8/5!
Look for common prime factors of the numerator and the denominator.
We can rewrite 5! as 5*4*3*2*1=5*2*2*3*2=2*2*2*3*5
Are there three 2s, one 3 and one 5 in the numerator?
Yes, there are.
Therefore all of those expressions will yield an integer.
I,II and III
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xcusemeplz2009
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for the first two there is no doubt , its an integer bcoz 6! and 8! will cancel out.
for the third one denominator is 120(5!=120) and 1n numerator 12*10 will cancel 120 so an integer
for the third one denominator is 120(5!=120) and 1n numerator 12*10 will cancel 120 so an integer
It does not matter how many times you get knocked down , but how many times you get up
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Scott@TargetTestPrep
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Before actually solving this problem, let's review how factorials can be expanded and expressed. As as example, we can use 5!.Xbond wrote: I. 12! / 6!
II. 12! / 8!
III. 12! / 7!5!
A) I only
B) II only
C) III only
D) I and II only
E) I, II, and III
5! could be expressed as:
5!
5 x 4!
5 x 4 x 3!
5 x 4 x 3 x 2!
5 x 4 x 3 x 2 x 1!
Understanding how this factorial expansion works will help us work our way through each answer choice, especially answer choices 1 and 2.
I. 12!/6!
Since we know that factorials can be expanded, we now know that:
12! = 12 x 11 x 10 x 9 x 8 x 7 x 6!
Plugging this in for answer choice 1, we have:
(12 x 11 x 10 x 9 x 8 x 7 x 6!)/6! = 12 x 11 x 10 x 9 x 8 x 7, which is an integer.
II. 12!/8!
Once again, since we know that factorials can be expanded, we now know that:
12! = 12 x 11 x 10 x 9 x 8!
Plugging this in for answer choice 2, we have:
(12 x 11 x 10 x 9 x 8!)/8! = 12 x 11 x 10 x 9, which is an integer.
III. 12!/(7!5!)
Once again, since we know that factorials can be expanded, we now know that:
12! = 12 x 11 x 10 x 9 x 8 x 7!
Plugging this in for answer choice 3 gives us:
(12 x 11 x 10 x 9 x 8 x 7!)/(7!5!)
(12 x 11 x 10 x 9 x 8)/(5 x 4 x 3 x 2 x 1)
(12 x 11 x 10 x 9 x 8)/(12 x 10 x 1)
11 x 9 x 8, which is an integer.
We see that the quantities in Roman numerals I, II and III are all integers.
Alternate solution:
For any positive integers m, n and p,
1) If m > n, then m!/n! is always an integer.
2) If m = n + p, then m!/(n!p!) is always an integer (which is in fact mCp).
From the above two facts, we see that all three quotients in the Roman numerals must be integers.
Answer: E
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