[email protected] wrote:What is the positive integer n?
(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-C

Target question: What is the value of positive integer n?

Statement 1: For every positive integer m, the product m(m + 1)(m + 2) ... (m + n) is divisible by 16
First notice that m, m+1, m+2, m+3 etc are CONSECUTIVE INTEGERS

There's a nice rule says: The product of k consecutive integers is divisible by k, k-1, k-2,...,2, and 1
So, for example, the product of any 5 consecutive integers will be divisible by 5, 4, 3, 2 and 1
NOTE: the product may be divisible by other numbers as well, but these divisors are guaranteed.

So, if the product of m, m+1, m+2 ... m+n is divisible by 16, there are many possible values of n. Consider these two conflicting cases:
Case a: n = 15. This means that (m)(m+1)(m+2)...(m+n) is the product of 16 consecutive integers. So, by the above rule, the product is definitely divisible by 16
Case b: n = 16. This means that (m)(m+1)(m+2)...(m+n) is the product of 17 consecutive integers. So, by the above rule, the product is definitely divisible by 16
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: n² - 9n + 20 = 0
Factor to get: (n - 4)(n - 5) = 0
So, n = 4 or n = 5
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 2 says that n = 4 or n = 5
However, although n = 4 satisfies statement 2, it does not necessarily satisfy statement 1.
If n = 4, then there are 5 consecutive integers in the product (m)(m+1)(m+2)...(m+n), and having 5 consecutive integers does not necessarily ensure that the product is divisible by 16.
For example (1)(2)(3)(4)(5) = 120, 120 is NOT divisible by 16.

So, if n ≠ 4, then n MUST EQUAL 5 [since statement 2 tells us that n equals EITHER 4 OR 5]
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer = C

Cheers,
Brent

I should point out, if n = 5, then (m)(m+1)(m+2)...(m+n) is the product of 6 consecutive integers, and the product of 6 consecutive integers IS ALWAYS divisible by 16. Here's why:
First recognize that 6 consecutive integers will always contain 3 consecutive odd integers and 3 consecutive EVEN integers.

Now take a look at some consecutive EVEN integers: 2, 4, 6, 8, 10, 12, 14, 16,...

Notice that every second EVEN integer is divisible by 4.
So, if we have 3 consecutive EVEN integers, then we can be certain that at least one of them is divisible by 4. Plus the other two are definitely divisible by 2 (since they are even).
In other words, we can rewrite any 3 consecutive EVEN integers as 2a, 2b, and 4c for some integers a, b and c.
As you can see the product of 2a, 2b, and 4c = 16abc, which means the 3 even consecutive integers are DEFINITELY divisible by 16.

This means that the product of ANY 6 consecutive integers will be divisible by 16

Cheers,
Brent