For every positive even integer n, the function h(n) is defined to be

Hi,

This question tends to trouble most Test Takers and it is definitely tougher than a typical GMAT question.

As a general rule, Quant questions are almost always based on a pattern of some kind (math formula, math rule, Number Property, etc.). If you can't immediately deduce a pattern, then you might have to "play around" a bit with the question to try to deduce what the pattern is. In the broad sense, it's critical thinking: here's a weird situation - what can I do to figure it out?

Based on the description of the function in the prompt, we can run some "TESTS" to try to figure things out….

The H(n) is the product of all the even integers from 2 to n, inclusive.

So….
H(4) = 2x4 = 8
The prime factors of 8 are (2)(2)(2)
If we do H(4) + 1 = 9, then the prime factors are (3)(3)
NOTICE how NONE of the prime factors of 8 are in 9? That's interesting….

H(6) = 2x4x6 = 48
The prime factors of 48 are (2)(2)(2)(2)(3)
If we do H(6) + 1 = 49, then the prime factors are (7)(7)
NOTICE how NONE of the prime factors of 48 are in 49? That's interesting….and probably a pattern, since it's happened TWICE NOW.

From here, I'd have to deduce that this pattern holds true. With H(100), I know that there are LOTS of primes that go in (the largest of which is 47, which can be "found" in 94). I have to assume that NONE of them will go into H(100) + 1. Thus, the smallest prime would have to be greater than 47. The question doesn't actually ask us for the exact prime number though - it just asked for the range that the prime would fall into.

Final Answer:

The takeaway from all of this is that you shouldn't be afraid to "play" with a question a bit. In the end, you don't have to be a brilliant mathematician to answer this question, but you're also not allowed to just stare at it either.

GMAT assassins aren't born, they're made,
Rich

Hi All,

I've been asked to post this solution here, so here's another way to handle this question:

For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of H(100) + 1, then p is:

1. between 2 and 10
2. between 10 and 20
3. between 20 and 30
4. between 30 and 40
5. greater than 40

The main idea behind this prompt is:

"The ONLY number that will divide into X and (X+1) is 1."

In other words, NONE of the factors of X will be factors of X+1, EXCEPT for the number 1.

Here are some examples:
X = 2
X+1 = 3
Factors of 2: 1 and 2
Factors of 3: 1 and 3
ONLY the number 1 is a factor of both.

X = 9
X+1 = 10
Factors of 9: 1, 3 and 9
Factors of 10: 1, 2, 5 and 10
ONLY the number 1 is a factor of both.
Etc.

Since the H(100) is (100)(98)(96)....(4)(2)....we can deduce....
1) This product will have LOTS of different factors
2) NONE of those factors will divide into H(100) + 1.

H(100) contains all of the primes from 2 through 47, inclusive (the 47 can be "found" in the "94"), so NONE of those will be in H(100) + 1. We don't even have to calculate which prime factor is smallest in H(100) + 1; we know that it MUST be a prime greater than 47....and there's only one answer that fits.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

As per the definition of the question

h(100) = 2 x 4 x 6 x 8 x 10 x 12 x 14 ... and so on...98 x 100 (Total 50 terms)

=> h(100) = 2^50 (1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10......and so on...x 48 x 49 x 50)

This means h(100) is multiple of all prime numbers between 1 and 50

therefore h(100)+1 will leave a remainder of 1 when divided by any prime number from 1 to 50

therefore, p, which is a factor of h(100)+1, will certainly be greater than a prime numbers greater than 50

Hence, "p" must be greater than 40 as per the following options

Answer: Option E

Hi,
without any calculations we can make out that h(n) will consist of products of all numbers till 50and 2^50, as it is even integers so 100/2..
therefore when we add 1 to this product , it will not be a multiple of any number till 50..
so the smallest prime factor has to be >50..
E..

lets solve it..
h(100)= 2*4*6...*100..
= 2*1*2*2*2*3...2*50= 2^50*1*2*...50..
h(100)+1= 2^50*1*2*...50 +1..
so prime factor>50..
ans E >40

NOTE:- Merging topic. Please type the first few letters as TOPIC name correctly
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Important Concept: If integer k is greater than 1, and k is a factor (divisor) of N, then k is not a divisor of N+1
For example, since 7 is a factor of 350, we know that 7 is not a factor of (350+1)
Similarly, since 8 is a factor of 312, we know that 8 is not a factor of 313

Now let’s examine h(100)
h(100) = (2)(4)(6)(8)….(96)(98)(100)
= (2x1)(2x2)(2x3)(2x4)....(2x48)(2x49)(2x50)
Factor out all of the 2's to get: h(100) = [2^50][(1)(2)(3)(4)….(48)(49)(50)]

Since 2 is in the product of h(100), we know that 2 is a factor of h(100), which means that 2 is not a factor of h(100)+1 (based on the above rule)

Similarly, since 3 is in the product of h(100), we know that 3 is a factor of h(100), which means that 3 is not a factor of h(100)+1 (based on the above rule)

Similarly, since 5 is in the product of h(100), we know that 5 is a factor of h(100), which means that 5 is not a factor of h(100)+1 (based on the above rule)

.
.
.
.
Similarly, since 47 is in the product of h(100), we know that 47 is a factor of h(100), which means that 47 is not a factor of h(100)+1 (based on the above rule)

So, we can see that none of the primes from 2 to 47 can be factors of h(100)+1, which means the smallest prime factor of h(100)+1 must be greater than 47.

Answer = E

Cheers,
Brent

Important Concept: If integer k is greater than 1, and k is a factor (divisor) of N, then k is not a divisor of N+1
For example, since 7 is a factor of 350, we know that 7 is not a factor of (350+1)
Similarly, since 8 is a factor of 312, we know that 8 is not a factor of 313

Now let’s examine h(100)
h(100) = (2)(4)(6)(8)….(96)(98)(100)
= (2x1)(2x2)(2x3)(2x4)....(2x48)(2x49)(2x50)
Factor out all of the 2's to get: h(100) = [2^50][(1)(2)(3)(4)….(48)(49)(50)]

Since 2 is in the product of h(100), we know that 2 is a factor of h(100), which means that 2 is not a factor of h(100)+1 (based on the above rule)

Similarly, since 3 is in the product of h(100), we know that 3 is a factor of h(100), which means that 3 is not a factor of h(100)+1 (based on the above rule)

Similarly, since 5 is in the product of h(100), we know that 5 is a factor of h(100), which means that 5 is not a factor of h(100)+1 (based on the above rule)

.
.
.
.
Similarly, since 47 is in the product of h(100), we know that 47 is a factor of h(100), which means that 47 is not a factor of h(100)+1 (based on the above rule)

So, we can see that none of the primes from 2 to 47 can be factors of h(100)+1, which means the smallest prime factor of h(100)+1 must be greater than 47.

Answer:

Cheers,
Brent

H(100) will be a multiple of all of the prime factors below 50 because it is the product of all of the primes below 50 multiplied by 2 (94 is 47 x 2, 62 is 31 x 2, etc. If h(100) is a multiple of each of these, then when we add 1 it will throw us off of being a multiple of all of these, so the greatest prime factor will have to be greater than 50, which is greater than 40.

Dear Bunuel, if the question changed to positive ODD integer, will the answer be the same?

I think answer choice A should be between 2 and 10.
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Yes. 49th odd number is 97, which is prime.
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We are given that h(n) is defined to be the product of all the even integers from 2 to n inclusive. For example, h(8) = 2 x 4 x 6 x 8.

We need to determine the smallest prime factor of h(100) + 1. Before determining the smallest prime factor of h(100) + 1, we must recognize that h(100) and h(100) + 1 are consecutive integers, and consecutive integers will never share the same prime factors.

Thus, h(100) and h(100) + 1 must have different prime factors. However, rather than determining all the prime factors of h(100), let’s determine the largest prime factor of h(100). Since h(100) is the product of the even integers from 2 to 100 inclusive, let’s find the largest prime number such that 2 times that prime number is less than 100.

That prime number is 47, since 2 x 47 = 94, which is less than 100. The next prime after 47 is 53, and 2 x 53 = 106, which is greater than 100.

Therefore, 47 is the largest prime number that is a factor of h(100). In fact, all prime numbers from 2 to 47 are included in the prime factorization of h(100). Since we have mentioned that h(100) + 1 will not have any of the prime factors of h(100), all the prime factors in h(100) + 1, including the smallest one, must be greater than 47. Looking at the answer choices, only choice E can be the correct answer.

Answer: E
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Hi Bunuel,

Thanks for the explanation! It was helpful.
I tried to apply the same logic with simple numbers. Consider we have 15. So if try to find out the least prime factor of 15 , we must take 14 and check for its prime factors. They are 2 and 7. So according to the above theory, will the least prime factor of 15 be > 7??? It should not! ( here, do we need to check >2 or >7 ??)

Where am I going wrong??

Thanks,
Uma

Two consecutive numbers are co-prime, so 14 and 15 are also co-prime, they do not share any common factor but 1. The factors of 14 are 1, 2, 7, and 14 and the factors of 15 are 1, 3, 5, and 15. As you can see no common factors but 1 (including no common primes).

The reasons you are confused is that \(h(100)=2^{50}*50!\) has ALL prime numbers from 1 to 50 as its factors, thus \(h(100)+1=2^{50}*50!+1\) won't have ANY prime factor from 1 to 50.

But 14 does NOT have ALL primes from 1 to 14, so 15 might have some primes from 1 to 14 (but 14 won't have the same primes as 15).
_________________

Seems to me your argument is about order of operations in a sentence. I think addition would take place BEFORE a verbal noun modifier (prepositional phrase “of...”)
Therefore the {} brackets are redundant.

Posted from my mobile device

Solution:

Given h(n)=2*3*4*...n

h(100) + 1 =2*4*6....100 + 1

= 2 * 2*2*(2*3)*(2*4)*...(2*50) + 1

= 2^50 (1*2*3*...50) +1

= 2^50 (50!)+1

Given p is the smallest prime factor of h(100) + 1

h(100) and h(100) + 1 are co prime numbers as they are consecutive integers.

h(100) has 50! as its factor or all the prime numbers from 1-50 are its factors.

=>h(100)+1 will NOT have any prime number as factor from 1-50

=>For p>1 the smallest number must be >50 (option e)

Devmitra Sen(GMAT SME)
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Hey Bunuel chetan2u I liked solving this question and I'd like to solve more of these types - factorial, prime, and divisibility combined. Can you help me with a name of these types so that I can search more of these?

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Simple solution:

H(100)
= 2 * 4 * 6 ... * 98 * 100
= (2 to the power of 50)(1 * 2 * 3 ...* 48 * 49 * 50)

If we start looking at prime numbers :
2 is a factor of H(100)
2 cannot be a factor of H(100)+1

3 is a factor of H(100)
3 cannot be a factor of H(100)+1

5 is a factor of H(100)
5 cannot be a factor of H(100)+1

We can see that every prime number less than 50 is a factor of H(100)
In other words, H(100) is a multiple of every prime number less than 50

For any of these prime numbers (call it p), the NEXT multiple will be H(100)+p
So H(100) + 1 cannot be a multiple of p

Posted from my mobile device
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Hi Bunuel,
Wouldn't 2*4*6*8*10*..........98*100 be equal to 2(1*2*3*4*5*.......49*50) i.e. 2*50! and not 2^50*50!.
Rgds

Had the mathematical operation been ‘addition’, you would be correct.
For example
\(2+4+6=2*1+2*2+2*3=2(1+3)\)
\(2*4*6=(2*1)*(2*2)*(2*3)=2*1*2*2*2*3=(2*2*2)*(1*2*3)=2^3(1*2*3)\)
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