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What is the greatest prime factor of 68−38 ?
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18 Jun 2021, 01:36
18 Jun 2021, 01:36
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66% (01:18) correct
33% (01:44) wrong based on 24 sessions
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What is the greatest prime factor of \(6^8−3^8\) ?
A) 3
B) 11
C) 17
D) 19
E) 31
A) 3
B) 11
C) 17
D) 19
E) 31
Re: What is the greatest prime factor of 68−38 ?
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18 Jun 2021, 01:57
18 Jun 2021, 01:57
2
\(6^8− 3^8\)
\(3^8 (2^8 -1)\)
\(3^8 * 255\)
\(3^8 * 5 * 3 *17 \)
so greatest prime factor is 17
\(3^8 (2^8 -1)\)
\(3^8 * 255\)
\(3^8 * 5 * 3 *17 \)
so greatest prime factor is 17
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Re: What is the greatest prime factor of 68−38 ?
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18 Jun 2021, 05:13
18 Jun 2021, 05:13
1
Carcass wrote:
What is the greatest prime factor of \(6^8−3^8\) ?
A) 3
B) 11
C) 17
D) 19
E) 31
A) 3
B) 11
C) 17
D) 19
E) 31
6⁸ − 3⁸ is a DIFFERENCE OF SQUARES. So we can factor it.
6⁸ − 3⁸ = (6⁴ + 3⁴)(6⁴ - 3⁴)
= (6⁴ + 3⁴)(6² + 3²)(6² - 3²)
= (6⁴ + 3⁴)(6² + 3²)(6 + 3)(6 - 3)
= (6⁴ + 3⁴)(45)(9)(3)
= (6⁴ + 3⁴)(3)(3)(5)(3)(3)(3)
Hmmmm, we can see that the correct answer is "hiding" in the first number (6⁴ + 3⁴)
Let's factor out the 3⁴, to get:
6⁴ + 3⁴ = 3⁴(2⁴ + 1)
= 3⁴(16 + 1)
= 3⁴(17)
= (3)(3)(3)(3)(17)
So, 6⁸ − 3⁸ = (3)(3)(3)(3)(17)(3)(3)(5)(3)(3)(3)
So the correct answer is C
Cheers,
Brent
