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Consecutive integer problem

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Consecutive integer problem

by Stockmoose16 » Mon Aug 04, 2008 1:48 pm
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.
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Re: Consecutive integer problem

by beeparoo » Mon Aug 04, 2008 2:25 pm
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...
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Re: Consecutive integer problem

by Ian Stewart » Mon Aug 04, 2008 2:34 pm
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.
You're absolutely right. If you multiply five consecutive even integers:

-all of them must be even, and therefore we know the product is divisible by 2^5;
-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);
-at least two of them must not only be divisible by 2, but also by 4, since every second even number is divisible by 4, and one of these must actually be divisible by 8, since every fourth even number is divisible by 8.

Put all of that together, and the product must be divisible by (2^8)*3*5, or 3840 (which is, not coincidentally, equal to 2*4*6*8*10, the product of the five smallest positive even numbers, so we can't do any better than this). This number has 36 different divisors, so while they've only listed six, there are thirty correct answers to the question missing from their list.
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Re: Consecutive integer problem

by Stockmoose16 » Mon Aug 04, 2008 2:37 pm
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?
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Re: Consecutive integer problem

by Ian Stewart » Mon Aug 04, 2008 3:13 pm
Stockmoose16 wrote: 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?
You're both entirely correct. The question says 'list six factors of the product...', and the book happens to have listed the six most obvious factors. Had the question said 'list all of the numbers which must be factors...' then they'd have the wrong answer- 3, 5, 6, and a lot of other numbers must also be factors of the product of five consecutive even integers. I think beeparoo was pointing out that their answer was correct to the question in the way they have asked it, but you are also correct to point out that there are many other numbers we can be certain are factors (as I explained in my post above).

edited- I had written 'consecutive integers' instead of 'consecutive even integers'...
Last edited by Ian Stewart on Mon Aug 04, 2008 3:48 pm, edited 1 time in total.
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Re: Consecutive integer problem

by beeparoo » Mon Aug 04, 2008 3:16 pm
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?
Alright alright, "rest assured" is too limiting to be applied to this problem. As Ian Stewart has explained it:
-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);
All I am saying is that the authors never state that "6 is not a factor". They merely neglect to talk about it, like a bad date.

Perhaps you would feel rest assured if I simply wrote, "this explanation sux - it's ambiguous and misleading".

Sandra
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