Is n ! n! the largest factor of x = a ( a + 1 ) ( a + 2 ) × × ( a + n ) x = a(a+1)(a+2)\times\cdots\times(a+n) ?

Consider x = a ( a + 1 ) ( a + 2 ) × × ( a + n ) x = a(a+1)(a+2)\times\cdots\times(a+n) .

True or False?

The largest known factor of x x is n ! n! for a > 0 a > 0 .

Clarifications: The values of x x and a a are unknown. The value of n n is known. (In other words: Find the largest expression only including n n and constants, that must be a factor of x, no matter which value a a (and x x ) has.)

Notation: ! ! denotes the factorial notation. For example: 8 ! = 1 × 2 × 3 × × 8 8! = 1\times 2 \times 3 \times \cdots \times 8 .

Hint: Substitute low values for a a and n n to develop an intuition.

If you get this problem wrong, please see the solutions. Recommended Problems:

False True

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1 solution

Ron Lauterbach
Sep 24, 2017

The largest known factor is ( n + 1 ) ! (n+1)! , as there are n + 1 n+1 terms (don't forget a a ).

( n + 1 ) ! > n ! (n+1)!>n!

The correct answer is F a l s e \boxed{False} .

Shouldn't the largest factor is " a ( a + 1 ) ( a + 2 ) × × ( a + n ) a(a+1)(a+2)\times\cdots\times(a+n) " itself?

Pi Han Goh - 3 years, 8 months ago

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Sorry misunderstood you.

Ron Lauterbach - 3 years, 8 months ago

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The largest factor of any integer N N is N |N| itself.

E.g: All the positive factors of 12 are { 1 , 2 , 3 , 4 , 6 , 12 } \{1,2,3,4,6,12\} , so the largest factor of 12 is 12 itself.

Pi Han Goh - 3 years, 8 months ago

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@Pi Han Goh Assumptions: The value of x x and a a are unknown. The value of n n is known.

Ron Lauterbach - 3 years, 8 months ago

@Pi Han Goh True or False? The largest known factor of x x is n ! n! for a > 0 a > 0 .

Ron Lauterbach - 3 years, 8 months ago

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@Ron Lauterbach Yes, I read that. I still don't think that is clear.

Pi Han Goh - 3 years, 8 months ago

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@Pi Han Goh Does this change clarify everything?

Clarifications: The values of x x and a a are unknown. The value of n n is known. In the question known refers to knowing an exact value and not an expression with unknown variables.

Ron Lauterbach - 3 years, 8 months ago

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@Ron Lauterbach Okay, I think this is the best possible way to phrase this:

( n + 1 ) ! 0 ! , ( n + 2 ) ! 1 ! , ( n + 3 ) ! 2 ! , \dfrac{(n+1)!}{0!} , \dfrac{(n+2)!}{1!} , \dfrac{(n+3)!}{2!} , \ldots

In terms of n n , what is the greatest common divisor of all the numbers above?

Pi Han Goh - 3 years, 8 months ago

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@Pi Han Goh This is kind of unrelated to my problem.

Ron Lauterbach - 3 years, 8 months ago

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@Ron Lauterbach First of all my problem goes from 1 to n and not from n+1 to infinity.

Ron Lauterbach - 3 years, 8 months ago

@Pi Han Goh For now, I made these changes:

Clarifications: The values of x x and a a are unknown. The value of n n is known. (In other words: Find the largest expression only including n n and constants, that must be a factor of x, no matter which value a a (and x x ) has.)

Would you say, this is clear enough without having to rephrase the entire problem?

Ron Lauterbach - 3 years, 8 months ago

It says the largest known factor. I specified, that we don't know a (and even x to clarify).

Ron Lauterbach - 3 years, 8 months ago

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Ah I see. Then I recommend that you change your question to:

Let x a = a ( a + 1 ) ( a + 2 ) ( a + n ) x_a = a(a+1)(a+2) \cdots(a+n) for positive integers a a and n n .

True or False?

The greatest common divisor of x a , x a + 1 , x a + 2 , x_a, x_{a+1} , x_{a+2} , \ldots is n ! n! .

Pi Han Goh - 3 years, 8 months ago

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@Pi Han Goh The answer to the question is still valid, even if the question is misunderstood, isn't it?

Ron Lauterbach - 3 years, 8 months ago

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@Ron Lauterbach Yes, this is a good question, but it's just hard to understand what you meant.

Pi Han Goh - 3 years, 8 months ago

@Pi Han Goh How do you create these boxes?

Ron Lauterbach - 3 years, 8 months ago

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@Ron Lauterbach Write like this:

> This is a box.

Pi Han Goh - 3 years, 8 months ago

@Pi Han Goh In your example it is unclear what x a , x a + 1 , x a + 2 , x_a, x_{a+1} , x_{a+2} , \ldots refers to.

Ron Lauterbach - 3 years, 8 months ago

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@Ron Lauterbach I've already written the definition of x a x_a .

Pi Han Goh - 3 years, 8 months ago

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@Pi Han Goh Yes, but writing x a x_a ( x a + 1 (x_{a+1} , ... is confusing (in general).

Ron Lauterbach - 3 years, 8 months ago

I didn't change anything so far. What I said previously (now deleted) was referring to a typo (It said + not × \times .

Ron Lauterbach - 3 years, 8 months ago

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