D01-20

Official Solution:


If \(a\), \(b\), and \(c\) are integers and \(a < b < c\), are \(a\), \(b\), and \(c\) consecutive integers?

Note that:

(1) The median of \((a!, \ b!, \ c!)\) is an odd number.

This implies that \(b!\) is odd. Therefore, \(b\) must be either 0 or 1. However, if \(b = 0\), then \(a\) would be a negative number (since \(a < b\)), making \(a!\) undefined, which is not possible. Therefore, \(a = 0\) and \(b = 1\) (yielding \((a!, \ b!, \ c!)=(1, 1, c!)\)). If \(c = 2\), the answer is YES, but if \(c\) is any other number, the answer is NO. Not sufficient.

(2) \(c!\) is a prime number.

This implies that \(c=2\). Not sufficient.

(1) + (2) From the previous analysis, we know that \(a = 0\), \(b = 1\), and \(c = 2\). Thus, the answer to the question is YES. Sufficient.


Answer: C
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That's NOT correct. For (2) we don't know whether a and b are positive integers. WE cannot use the info from (1) when considering the second statement.
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I got this question wrong in the test, because I didn't pay special attention to every detail given in the question.

Here's the right solution :

1. The median of a! b! c! is an odd number, it means that b is either 0 or 1. Since we can have a!, it means that a is greater than or equal to 0. We also know that a < b < c, so a = 0, b = 1, but we have no information about c. SO, this is insufficient.

2. c! is a prime number, this means that c = 2. Here we have no information about a and b, so Insufficient.

Combining 1 & 2, we get that a = 0, b = 1, c=2.

So, C

I think this question is poor and helpful.
I believe the given answer is incorrect and the correct one is B -> statement 2 by itself is sufficient. Indeed, if c! is prime then it must be 2 and since a < b < c and a,b,c are INTEGERS AND factorial is not defined for negative numbers then a and b have no other choice but to be 0 and 1 respectively! Thus, statement 2 is enough by itself. I would like to hear your comments. Thanks.

Bunuel and others:
I have similar doubt like others. I appreciate your inputs.

Lets ignore Stmt A for now.
Question says, If a, b, and c are integers and a<b<c, are a, b, and c consecutive integers?
Stmt(1): ignore.
Stmt(2): c! is a prime number.
only possible number is 2. !2 =2 a prime number.
Now, a<b<c and they are integers. The only possible combination i see here is 0,1,2. I ignored combination with negative numbers such as -1,1,2 because !negative is undefined.
Where am i going wrong here?

You do not have factorial of a or b in stem or in (2) so you cannot ignore negative numbers.
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Bunuel
As per the solution given:
(1) The median of {a!, b!, c!} is an odd number. This implies that b!=odd. Thus b is 0 or 1. But if b=0, then a is a negative number, so in this case a! is not defined. Therefore a=0 and b=1, so the set is {0!, 1!, c!}={1, 1, c!}. Now, if c=2, then the answer is YES but if c is any other number then the answer is NO. Not sufficient.

it says b is 0 or 1. Is 0 an odd number?

0 is an even integer. But 0! = 1 = odd.
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How is 0 factorial 1?

0! = 1.

It's a math property you need not know why it's so but if still interested check this video:


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How can we determine that c must be 2 (not any other prime number) from the :

2) c! is a prime number.

n! = 1*2*3*...*n

Now, if c is any other number than 2, c! will have at least 2 and 3 as its factor and we know that a prime number has only two factors 1 and itself, thus for c! to be prime c must be 2. For example, if c = 4, then c! = 4! = 1*2*3*4, which is not a prime.

Hope it's clear.
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I think this is a high-quality question and I agree with explanation. Great question! Tests a mix of multiple basic facts on numbers.

Bunuel Would you be able to help me with this?

How do we know that "The median of {a!, b!, c!} is an odd number" means that b is odd?

The important point to remember is that factorial of negative number is undefined. Option A says "The median of {a!, b!, c!} is an odd number" , which means b! is odd.

Only 2 options results in odd numbers either 0! or 1!. But since a < b, b cannot be 0 (which means a < b , and a is negative and a! is undefined) , so b is 1 (odd number).

I hope i answered your question.
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I think this is a high-quality question and I agree with explanation.

I think this is a high-quality question and I agree with explanation.

I don't agree with the explanation. What if C=3. It is a prime number and it is greater than 1. The answer to the question will be no in this case. Would not option E make more sense?

(2) says that c! is prime. If c = 3, then c! = 6, which is not a prime. So, c cannot be 3.

Does this make sense?
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