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Lisa [10]
3 years ago
7

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, inclusiv

e. If p is the smallest prime factor of h(100) + 1, then p is ____.
Mathematics
1 answer:
Ugo [173]3 years ago
7 0

Answer:

More than 50

Step-by-step explanation:

To solve, we need to first see that the function is h(n). Picking main points from the question statement:

  • h(n) is the product of all even integers (From 2 to n)
  • p is the smallest factor of h(100)+1
  • h(100)+1 , here n=100

From here, we can write h(100) as:

h(100) = 2 * 4 * 6 * 8 * ...... * 100

h(100) = 2^{50} * (1*2*3*......*50)= 2^{50} * 50!

so,

h(100)+1 =(2^{50} * 50! )+1

Now two numbers,

h(100) and h(100)+1 are consecutive integers and since they are consecutive so they are co-prime. Hence they only have common factor of 1. Example, 13 and 14 have only common factor of 1

As h(100) has all prime numbers from 1 to 50 and according to above statement h(100)+1 won't have any prime factor from 1 to 50, so the smallest prime factor p is greater than 50.

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Please help i dont understand what its asking
Arisa [49]

Multiply both the numerator and denominator by 2+\sqrt3, which is called the "conjugate" of 2-\sqrt3:

\dfrac{5+\sqrt3}{2-\sqrt3}\cdot\dfrac{2+\sqrt3}{2+\sqrt3}

Why do we pick this number? Recall the difference of squares factorization:

a^2-b^2=(a-b)(a+b)

Now replace a=2 and b=\sqrt3. Then

(2-\sqrt3)(2+\sqrt3)=2^2-(\sqrt3)^2=4-3=1

So in your fraction, you end up with

\dfrac{5+\sqrt3}{2-\sqrt3}=\dfrac{(5+\sqrt3)(2+\sqrt3)}1=(5+\sqrt3)(2+\sqrt3)

Finally, just expand the product.

(5+\sqrt3)(2+\sqrt3)=10+7\sqrt3+(\sqrt3)^2=\boxed{13+7\sqrt3}

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