Problem Solving

Stockmoose16 wrote:This is from the MGMAT strategy book:

List six factors of the product of 5 consecutive even integers

The answer is: 1, 2, 4, 8, 16, 32

**The explanation says that each consecutive even integer has a factor of 2, So any combination of (2*2*2*2*2) would be a factor, as well as 1. But here's my issue: shouldn't 6 also be a factor? Here's my logic, using the consecutive even integers 2, 4, 6, 8, 10, as an example.

The prime factor of 2 is 2
The prime factors of 4 are 2,2
The prime factors of 6 are 3, 2
The prime factors of 8 are 2, 2, 2
The prime factors of 10 are 2, 5

So why wouldn't 3*2 (6) be one of the products of 5 consecutive even integers? And, for that matter, why wouldn't 5*2 (10) be one, too? I'm so confused.

beeparoo wrote:

Stockmoose16 wrote:This is from the MGMAT strategy book:

List six factors of the product of 5 consecutive even integers

The answer is: 1, 2, 4, 8, 16, 32

**The explanation says that each consecutive even integer has a factor of 2, So any combination of (2*2*2*2*2) would be a factor, as well as 1. But here's my issue: shouldn't 6 also be a factor? Here's my logic, using the consecutive even integers 2, 4, 6, 8, 10, as an example.

The prime factor of 2 is 2
The prime factors of 4 are 2,2
The prime factors of 6 are 3, 2
The prime factors of 8 are 2, 2, 2
The prime factors of 10 are 2, 5

So why wouldn't 3*2 (6) be one of the products of 5 consecutive even integers? And, for that matter, why wouldn't 5*2 (10) be one, too? I'm so confused.

Um, I'm in agreement with you that 6 is also a potential factor. BUT I don't think that the authors have discounted the possibility for "6" or even "10" just because they haven't listed it.

The numbers they have listed are those that are guaranteed to be factors of "the product of 5 consecutive even integers" because they are powers of 2 (up to 2^5).

The 5 consecutive even integers can be identified as:
2(n)
2(n + 1)
2(n + 2)
2(n + 3)
2(n + 4)

Regardless of what "n" may be (i.e. n = 1 000 001), you can be rest assured that the factors will always include:
1, 2, 2^2, 2^3, 2^4, and 2^5

Hope this helps...

Stockmoose16 wrote:

beeparoo wrote:

Stockmoose16 wrote:This is from the MGMAT strategy book:

List six factors of the product of 5 consecutive even integers

The answer is: 1, 2, 4, 8, 16, 32

**The explanation says that each consecutive even integer has a factor of 2, So any combination of (2*2*2*2*2) would be a factor, as well as 1. But here's my issue: shouldn't 6 also be a factor? Here's my logic, using the consecutive even integers 2, 4, 6, 8, 10, as an example.

The prime factor of 2 is 2
The prime factors of 4 are 2,2
The prime factors of 6 are 3, 2
The prime factors of 8 are 2, 2, 2
The prime factors of 10 are 2, 5

So why wouldn't 3*2 (6) be one of the products of 5 consecutive even integers? And, for that matter, why wouldn't 5*2 (10) be one, too? I'm so confused.

Um, I'm in agreement with you that 6 is also a potential factor. BUT I don't think that the authors have discounted the possibility for "6" or even "10" just because they haven't listed it.

The numbers they have listed are those that are guaranteed to be factors of "the product of 5 consecutive even integers" because they are powers of 2 (up to 2^5).

The 5 consecutive even integers can be identified as:
2(n)
2(n + 1)
2(n + 2)
2(n + 3)
2(n + 4)

Regardless of what "n" may be (i.e. n = 1 000 001), you can be rest assured that the factors will always include:
1, 2, 2^2, 2^3, 2^4, and 2^5

Hope this helps...

Why can't you be "rest assured" that 6 is a factor? Would there ever be a case where the product of five consecutive even integers wouldn't be divisible by 3? Would there ever be a case where the product of 5 consecutive even integers wouldn't be divisible by 5?

-at least one of them must be divisible by 3 (every third even integer is, so one of them must be);
-at least one of them must be divisible by 5 (every fifth even integer is, so one of them must be);