What is the positive integer n?

Question

(1) For every positive integer m, the product m(m + 1)(m + 2) ... (m + n) is divisible by 16
(2) n^2 - 9n + 20 = 0

Answer 1

Is this problem looking for one answer that is true for both?  If so, the answer is n=5.

 

The second part of the problem is factoring, were looking for two numbers that add to equal -9 and multiply to equal 20.  So it factors as (n-4)(n-5), n=4 and n=5

 

The first part of the problem is a harder, the second answer helped find the first.  Since this has to work with every positive number m, start with m=1.  So our first possible products would be:

(1)(1+1)(1+2)...(1+n)

or

(1)(2)(3)...(1+n)

Continue writing consecutive numbers until you get enough terms with 2 in them to get 2^4 or 16.

(1)(2)(3)(4)(5)(6) = (1)(2)(3)(2*2)(5)(2*3)

This has 4 2s as factors, so 16 can be factored out.  Any increase in m would maintain at least 4 2s as factors.

Answer 2

Let's start with statement 2 because it looks easier.

We can rewrite this as (n - 4)*(n-5) = 0

n could be 4 or 5

NOT SUFFICIENT

 

For statement 1 let's test cases using smart numbers.  m is an integer greater than 0.  Use 1.

1(1+1)*(1+2)*...(1+n)/16 is an integer

1*2*3*...*(n+1)/16 is an integer

Prime factor 16 to rewrite as . . .

1*2*3*...*(n+1)/2^4 is an integer.  So the numerator just needs at least four factors of 2 for the denominator to cancel.

Too many possibilities for n

NOT SUFFICIENT

 

Together

n=4 doesn't work because there are only three 2's in the numerator

n=5 works because there are now four 2's in the numerator

SUFFICIENT