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# For each positive integer k, let ak = (1 + 1/(k+1)). Is the product a1

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Re: For each positive integer k, let ak = (1 + 1/(k+1)). Is the product a1 [#permalink]
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gmatt1476 wrote:
For each positive integer k, let $$a_k = (1 + \frac{1}{k+1})$$. Is the product $$a_1a_2 … a_n$$ an integer?

(1) n + 1 is a multiple of 3.
(2) n is a multiple of 2.

DS59851.01

The correct answer is B. The detailed working can be found in the attached document.
Attachments

My Solution.doc [40 KiB]

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Re: For each positive integer k, let ak = (1 + 1/(k+1)). Is the product a1 [#permalink]
gmatt1476 wrote:
For each positive integer k, let $$a_k = (1 + \frac{1}{k+1})$$. Is the product $$a_1a_2 … a_n$$ an integer?

(1) n + 1 is a multiple of 3.
(2) n is a multiple of 2.

DS59851.01
coylahood wrote:
gmatt1476 wrote:
For each positive integer k, let $$a_k = (1 + \frac{1}{k+1})$$. Is the product $$a_1a_2 … a_n$$ an integer?

(1) n + 1 is a multiple of 3.
(2) n is a multiple of 2.

DS59851.01

The correct answer is B. The detailed working can be found in the attached document.

Sorry I realised a mistake with my detailed solution. I have uploaded the corrected version.
Attachments

Updated Solution.doc [41 KiB]

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Re: For each positive integer k, let ak = (1 + 1/(k+1)). Is the product a1 [#permalink]
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When faced with sequences, mostly it makes sence to list some values and grasp the pattern.

In this case it will start with 3/2 and each integer after that will increase both numerator and denominator by 1.

A1= 3/2
A2=4/3
A3=5/4
A4=6/5
A5=7/6

Now the Q is whether the multiplication of all those up to N will be an integer. List up the results you just established nex to each other as if you were multiplying them. You will notice that the numerators and denominators ALWAYS cancel out.

3/2 x 4/3 x 5/4 x 6/5 x 7/6 ......

The only one that doesn´t cancel out is the Denominator from the first term (A1=3/2, so 2 will always stay down there). This means that whether multiplying all terms ends up as an integer depends on whether the numerator from the last term is EVEN or ODD (since an even number divided by 2 will always be an integer and an odd number divided by 2 will never be an integer).

Now, since the numerators alternate between even and odd, we only need to know whether the last term is an even or an odd number.

(1) n + 1 is a multiple of 3.

Understand here that any number (whether odd or even) will have a consecutive integer that can be odd or even. There is no pattern such that if n= even, n+1 is always/never divisible by three.

Eg.: n=2, n+1=3 (multiple of 3). n=5, n+1=6 (also multiple of three)
As you can see we can get an odd and even number for n, so this doesn´t help us.

(2) n is a multiple of 2.

This means that n is even (since any number multiplied by an even number - 2 is an even number - will be even).
Exactly what we needed.

I hope it helps! Let the Kudos button burn :D
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Re: For each positive integer k, let ak = (1 + 1/(k+1)). Is the product a1 [#permalink]
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coylahood wrote:
gmatt1476 wrote:
For each positive integer k, let $$a_k = (1 + \frac{1}{k+1})$$. Is the product $$a_1a_2 … a_n$$ an integer?

(1) n + 1 is a multiple of 3.
(2) n is a multiple of 2.

DS59851.01

The correct answer is B. The detailed working can be found in the attached document.

The sequence follows a pattern as follows,
Element-n = (n+2)/(n+1)
Element = {3/2 , 4/3, 5/4, 6/5, 7/6, 8/7 .... }

Evaluating, (1) n + 1 is a multiple of 3.
possible values of n = {2, 5, 8, 11 ...}
if n=2, we get prod = 3/2*4/3 = integer (INT-YES)
if n=5, we get prod = 3/2*4/3*5/4*6/5*7/6 = 7/2 = fraction (INT-NO)
Not Sufficient

Evaluating, (2) n is a multiple of 2.
for all multiples of the prod will be integer.
every position at multiple of 2 will have numerator as multiple of two, which will cancel with the denominator 2 in (3/2)[first element]
Sufficient!
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Re: For each positive integer k, let ak = (1 + 1/(k+1)). Is the product a1 [#permalink]
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This sequence follows a pattern: $$\frac{3}{2} * \frac{4}{3} * \frac{5}{4} * \frac{6}{5} * \frac{7}{6}$$...

In the first two terms, the 3 on each term cancels out. As we add more terms we see the same pattern in each of the terms. The last term needs to be even in order for the term to be divisible by the 2 in the first term.

Therefore the question rephrased: Is the last term even?

(1) We can't say for sure.

6 - 1 = odd
3 - 2 = even

INSUFFICIENT.

(2) Gives us our answer. SUFFICIENT.

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Re: For each positive integer k, let ak = (1 + 1/(k+1)). Is the product a1 [#permalink]
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gmatt1476 wrote:
For each positive integer k, let $$a_k = (1 + \frac{1}{k+1})$$. Is the product $$a_1a_2 … a_n$$ an integer?

(1) n + 1 is a multiple of 3.
(2) n is a multiple of 2.
DS59851.01

Let's plug in a couple of values of k and check for integer/non-integer outcome.

1. $$k = 1$$ yields $$1 + \frac{1}{2} = \frac{3}{2}$$
2. $$k = 2$$ yields $$1 + \frac{1}{3} = \frac{4}{3}$$
3. $$k = 3$$ yields $$1 + \frac{1}{4} = \frac{5}{4}$$

Multiplying the above three values $$= \frac{3}{2} * \frac{4}{3} * \frac{5}{4} = \frac{5}{2}$$ not an integer, but observe that multiplying the first two values yields an integer $$\frac{3}{2} * \frac{4}{3} = 2$$. You will observe the same pattern if you try some more values.

To summarize if $$n$$ is even then the outcome is an integer and if $$n$$ is odd then the outcome is not an integer.

Option-A: Insufficient

1. $$n = 8$$ then $$n + 1$$ is a multiple of $$3$$ and the product is an integer
2. $$n = 5$$ then $$n + 1$$ is a multiple of $$3$$ but the product is not an integer

Option-B: Sufficient

$$n$$ is a multiple of $$2$$ which means that n is even and as we have proved at the start that the outcome will be an integer

Ans B
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Re: For each positive integer k, let ak = (1 + 1/(k+1)). Is the product a1 [#permalink]
For questions like this one Ive found that a good approach is to always list the first 3-4 outputs and see what happens.

a1 = 1,5
a2 = 1,33
a3 = 1,25
a4 = 1,2

a1 = 1,5
a1*a2 = 1,5*1,33 = 2
a1*a2*a3 = 2*1,25 = 2,5
a1*a2*a3*a4 = 2,5*1,2 = 3

After only four products we can discern the pattern. Every second product seems to be an integer. Or in other words, when n is even, the product is an integer.

(1) With this information, n may or may not be even.
(2) With this information, n must be even.
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Re: For each positive integer k, let ak = (1 + 1/(k+1)). Is the product a1 [#permalink]
The most important first step here is to write out the first several factors of the product, in improper fraction form:
3/2 * 4/3 * 5/4 * 6/5 * 7/6 * 8/7 * 9/8 * 10/9 * ... * (n+2)/(n+1)

How did I know I should use improper fractions, as opposed to mixed fractions (e.g. 1 1/2) or decimals (e.g. 1.5)?

For one thing, I can't think of a single GMAT problem I've ever seen in which mixed fractions were useful, and decimals are very rarely useful. In fact, I've had problems in which the question stem and the answer choices were all in decimal form, and yet I converted the decimals from the question stem into fractions, did the work, and then converted my answer back to decimal form at the end to match the answer choices.

In this particular problem, the word "product" is a big red flag, warning us to avoid using mixed fractions or decimals.

Now, going back to the question: we want to know whether the product is an integer. So, I have to look at the product and wonder: under what circumstances would this product, which includes a bunch of consecutive integers in the numerators and a bunch of consecutive integers in the denominator, be an integer - and under what circumstances would it not be an integer.

Just to rephrase that product in a more convenient form:
(3*4*5*6*7*8*9*10*...*(n+2)) / (2*3*4*5*6*7*8*9*...*(n+1))

Why would that EVER not be an integer??

Well, the denominator starts from 2, which the numerator is missing.
Thankfully, the numerator does have one extra factor at the end, (n+2).

So, if (n+2) contains a '2' in its prime box, it will compensate for that missing '2' at the beginning, giving us a YES answer to the question.
But, if (n+2) doesn't contain a '2' in its prime box (i.e. it's odd), then the answer to the question is NO.

So, rephrase the question: is (n+2) even?
Since consecutive integers alternate between odd an even, a further rephrase could look like this:
Is n even?

Now I'm ready to evaluate the statements.
Start with statement 2, as it tells me precisely what I wanted to know! Eliminate choices ACE.
Statement 2 doesn't provide any information about divisibility by 2.

Since the divisors 2 and 3 share no common factors (they are coprime), information about divisibility by 3 sheds no light about divisibility by 2. Eliminate D and choose B.
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Re: For each positive integer k, let ak = (1 + 1/(k+1)). Is the product a1 [#permalink]
nick1816 wrote:
$$a_k = (1 + \frac{1}{k+1})$$
$$a_k = \frac{k+2}{k+1})$$

$$a_1a_2 … a_n$$= $$\frac{3}{2}*\frac{4}{3}*\frac{5}{4}*.........*\frac{n+2}{n+1}$$

$$a_1a_2 … a_n$$=$$\frac{n+2}{2}$$

$$a_1a_2 … a_n$$= $$\frac{n}{2}+1$$

$$\frac{n}{2}+1$$ is an integer, if $$\frac{n}{2}$$ is an integer

So basically question stem is whether n is a multiple of 2 or not.

Statement 1- n+1 is a multiple of 3

if n+1 is 3, n is 2 (even)
if n+1 is 6, n is 5 (odd)

Insufficient

Statement 2- n is multiple of 2 or even

Sufficient

gmatt1476 wrote:
For each positive integer k, let $$a_k = (1 + \frac{1}{k+1})$$. Is the product $$a_1a_2 … a_n$$ an integer?

(1) n + 1 is a multiple of 3.
(2) n is a multiple of 2.

DS59851.01

Thanks nick1816 . could you elaborate more on this to arrive at $$\frac{n+2}{2}$$ ? Thanks
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Re: For each positive integer k, let ak = (1 + 1/(k+1)). Is the product a1 [#permalink]
Kimberly77 wrote:
Thanks nick1816 . could you elaborate more on this to arrive at $$\frac{n+2}{2}$$ ? Thanks

Hello Kimberly77 ,

We have discussed this question on YouTube. However, since you require assistance with just a part of the solution, I will elaborate that here.

Let’s discuss how we arrived at a$$_1$$a$$_2$$….a$$_n$$ = $$\frac{(n + 2)}{2}$$.

The question statement tells us that any term a$$_k$$ is represented as a$$_k$$ = 1 + $$\frac{1}{k + 1}$$ where k is a positive integer.
We can simplify this expression of a$$_k$$ as follows:
• a$$_k$$ = 1 + $$\frac{1}{(k + 1)}$$
• = $$\frac{(k + 1 + 1)}{(k + 1)}$$ --- [Taking (k + 1) as the common denominator]
• = $$\frac{(k + 2)}{(k + 1)}$$ ----- (1)

Now, we need to find something about the product of the first ‘n’ terms of this sequence, that is, about a1a2….an. So, let’s first find this product in its simplest form
Putting k = 1 in (1), we get a$$_1$$ = $$\frac{(1 + 2)}{(1 + 1)}$$ = 3/2
Putting k = 2, we get a$$_2$$ = $$\frac{(2 + 2)}{(2 + 1)}$$ = 4/3
Putting k = 3, we get a$$_3$$ = $$\frac{(3 + 2)}{(3 + 1)}$$ = 5/4

Let’s skip forward and calculate the last 2 terms:
Putting k = n – 1, we get a$$_n$$$$_-$$$$_1$$ = $$\frac{(n – 1 + 2)}{(n – 1 + 1)}$$ = $$\frac{(n +1)}{n}$$
Putting k = n, we get a$$_n$$ = $$\frac{(n + 2)}{(n + 1)}$$
Therefore, a$$_1$$a$$_2$$a$$_3$$….a$$_n$$$$_-$$$$_1$$a$$_n$$ = $$\frac{3}{2}$$ * $$\frac{4}{3}$$ * $$\frac{5}{4}$$ *…. * $$\frac{(n + 1)}{n}$$ * $$\frac{(n + 2)}{(n + 1)}$$ ---- (2)

Now closely observe (2).

The numerator (3) of the first term (3/2) gets cancelled by the denominator (3) of the second term (4/3).
Similarly, the numerator (4) of second term (4/3) gets cancelled by the denominator (4) of the third term (5/4).
The pattern would continue till the very end. The numerator (n + 1) of a$$_n$$$$_-$$$$_1$$ (n + 1/n) gets cancelled by the denominator (n + 1) of an [(n + 2)/(n + 1)].

So, every term in the numerator and denominator gets cancelled EXCEPT the denominator of first term a$$_1$$: 2 and the numerator of the last term a$$_n$$: n + 2
Thus, we finally are left with (n + 2) in the numerator and 2 in the denominator.
Therefore, we get a$$_1$$a$$_2$$a$$_3$$….a$$_n$$ = $$\frac{(n + 2)}{2}$$

And done!
I hope you now understand this part and can use this to proceed further with the solution.

Best Regards,
Ashish Arora,
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