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fla162
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For every positive even integer n, the function h(n) is defined to be Updated on: 13 Oct 2015, 19:47
Originally posted by fla162 on 13 Oct 2015, 15:02.
Last edited by Bunuel on 13 Oct 2015, 19:47, edited 1 time in total.
RENAMED THE TOPIC.
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58% (01:24) correct 42% (01:54) wrong based on 326 sessions

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For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. For instance, h(10)= 2x4x6x8x10. What is the greatest prime factor of h(28)+h(30)?

a) 5
b) 7
c) 13
d) 29
e) 31
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E
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Marchewski
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For every positive even integer n, the function h(n) is defined to be 17 Oct 2015, 09:59
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\(h(28) + h(30) = h(28) + h(28) * 30 = h(28) * (1 + 30) = h(28) * 31\)

31 will be the greatest prime factor.
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Bunuel
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For every positive even integer n, the function h(n) is defined to be 13 Oct 2015, 19:54
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fla162
For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. For instance, h(10)= 2x4x6x8x10. What is the greatest prime factor of h(28)+h(30)?

a) 5
b) 7
c) 13
d) 29
e) 31
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\(h(28) = 2*4*6*...*28 = (2*1)*(2*2)*(2*3)*...*(2*14) = 2^{14}*14!\)

\(h(30) = 2*4*6*...*28*30 = (2*1)*(2*2)*(2*3)*...*(2*14)*(2*15) = 2^{15}*15!\)

\(h(28) + h(30) = 2^{14}*14! + 2^{15}*15! = 2^{14}*14!(1 + 2*15) = 2^{14}*14!*31\).

Answer: E.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to rule 3. Thank you.
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For every positive even integer n, the function h(n) is defined to be 15 Jun 2020, 17:47
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Bunuel
fla162
For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. For instance, h(10)= 2x4x6x8x10. What is the greatest prime factor of h(28)+h(30)?

a) 5
b) 7
c) 13
d) 29
e) 31
Show more

\(h(28) = 2*4*6*...*28 = (2*1)*(2*2)*(2*3)*...*(2*14) = 2^{14}*14!\)

\(h(30) = 2*4*6*...*28*30 = (2*1)*(2*2)*(2*3)*...*(2*14)*(2*15) = 2^{15}*15!\)

\(h(28) + h(30) = 2^{14}*14! + 2^{15}*15! = 2^{14}*14!(1 + 2*15) = 2^{14}*14!*31\).

Answer: E.


Could someone please explain or show how the terms are factored to arrive at the final simplifed expression? I know that factoring is happening with both the factorials and the exponents—but precisely how, step by step, are the exponent and factorial terms drawn (factored) out? I think I've worked through the process, but I don't think I'd feel confident attacking a similar problem. For example, why exactly does it become (1 + 2*15) and not (2 + 1*15). I'd very much like to see the steps so I can practice them. Thanks!
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For every positive even integer n, the function h(n) is defined to be 16 Jun 2020, 00:27
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hughmac13
Bunuel
fla162
For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. For instance, h(10)= 2x4x6x8x10. What is the greatest prime factor of h(28)+h(30)?

a) 5
b) 7
c) 13
d) 29
e) 31

\(h(28) = 2*4*6*...*28 = (2*1)*(2*2)*(2*3)*...*(2*14) = 2^{14}*14!\)

\(h(30) = 2*4*6*...*28*30 = (2*1)*(2*2)*(2*3)*...*(2*14)*(2*15) = 2^{15}*15!\)

\(h(28) + h(30) = 2^{14}*14! + 2^{15}*15! = 2^{14}*14!(1 + 2*15) = 2^{14}*14!*31\).

Answer: E.


Could someone please explain or show how the terms are factored to arrive at the final simplifed expression? I know that factoring is happening with both the factorials and the exponents—but precisely how, step by step, are the exponent and factorial terms drawn (factored) out? I think I've worked through the process, but I don't think I'd feel confident attacking a similar problem. For example, why exactly does it become (1 + 2*15) and not (2 + 1*15). I'd very much like to see the steps so I can practice them. Thanks!
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Here: \(h(28) =(2*1)*(2*2)*(2*3)*...*(2*14) = 2^{14}*14!\) we have fourteen 2's. When they are factored out we get: \(2^{14}(1*2*3*...*14)=2^{14}*14!\)

Here: \(h(30) = (2*1)*(2*2)*(2*3)*...*(2*14)*(2*15) = 2^{15}*15!\) we have fifteen 2's. When they are factored out we get: \(2^{15}(1*2*3*...*14*15)=2^{15}*15!\)

Here: \(2^{14}*14! + 2^{15}*15! = 2^{14}*14!(1 + 2*15) = 2^{14}*14!*31\), we factored out 2^14*14 out of \(2^{14}*14! + (2^{14}*14!)*(2*15)\) and got \(2^{14}*14!(1 + 2*15) \).

Hope it'd clear.
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For every positive even integer n, the function h(n) is defined to be 16 Jun 2020, 01:01
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H(30) IS H(28)*30
SO:
H(28) (1+30)
H(28)*31

OR you can find that h(30) has 15 items with mean of 16 and h(28) 15 iteams with mean of 15, so:

16*15+15*15= 15*(16+15)=15*31 so 31
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For every positive even integer n, the function h(n) is defined to be 18 Jun 2020, 14:30
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fla162
For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. For instance, h(10)= 2x4x6x8x10. What is the greatest prime factor of h(28)+h(30)?

a) 5
b) 7
c) 13
d) 29
e) 31
Show more

Solution:

Since h(30) contains all the factors of h(28) in addition to the factor 30, we can factor h(28) from the sum. That is,

h(28)+h(30) = h(28)[1 + 30] = h(28) x 31

Since all the prime factors of h(28) are less than 28 itself, we see that the greatest prime factor is 31.

Answer: E
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For every positive even integer n, the function h(n) is defined to be 05 Dec 2024, 15:12
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