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What is the sum of all distinct prime factors of 6^12*49^2- 2^10*3^11*

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What is the sum of all distinct prime factors of 6^12*49^2- 2^10*3^11*  [#permalink]

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New post 28 Jan 2019, 02:31
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

57% (01:45) correct 43% (01:22) wrong based on 51 sessions

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What is the sum of all distinct prime factors of 6^12*49^2- 2^10*3^11*  [#permalink]

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New post 28 Jan 2019, 02:42
1
Bunuel wrote:
What is the sum of all distinct prime factors of \(6^{12}*49^2- 2^{10}*3^{11}*7^4\)?

A. 5
B. 9
C. 10
D. 12
E. 23


IMO E

\(2^{12} * 3^{12} * 7^4 - 2^{10} * 3^{11} * 7^4\)

\(2^{10} * 3^{11} * 7^4 * [ 12-1]\)

2+3+7+11

23
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Re: What is the sum of all distinct prime factors of 6^12*49^2- 2^10*3^11*  [#permalink]

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New post 28 Jan 2019, 02:44
\(6^{12}*7^4 - 6^{10}*3*7^4\)

\(6^{10}*7^4 (6^2 - 3)\)

\(3*11*3^{10}*2^{10}*7^4\)

\(2^{10}*3^{11}*7^4*11\)

Sum of distinct prime factors is 2+3+7+11 = 23

OPTION: E
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Re: What is the sum of all distinct prime factors of 6^12*49^2- 2^10*3^11*  [#permalink]

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New post 28 Jan 2019, 03:52
Bunuel wrote:
What is the sum of all distinct prime factors of \(6^{12}*49^2- 2^{10}*3^{11}*7^4\)?

A. 5
B. 9
C. 10
D. 12^7
E. 23


\(3^{12} 2^{12} 7^4 - 2^{10}3^4\)

= \(3^{11}2^{10}7^4 ( 3^12^2* - 1)\)

=\(3^{11}2^{10}7^4*11\)

3 + 2 + 7 + 11 = 23.

E is the correct answer.
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Re: What is the sum of all distinct prime factors of 6^12*49^2- 2^10*3^11*  [#permalink]

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New post 28 Jan 2019, 03:57
Bunuel wrote:
What is the sum of all distinct prime factors of \(6^{12}*49^2- 2^{10}*3^{11}*7^4\)?

A. 5
B. 9
C. 10
D. 12
E. 23


upon solving expression we woud get
2^10*3^11*7^4 * 11

sum = 2+3+7+11
23 IMO E
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What is the sum of all distinct prime factors of 6^12*49^2- 2^10*3^11*  [#permalink]

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New post 13 Feb 2019, 18:24
Bunuel wrote:
What is the sum of all distinct prime factors of \(6^{12}*49^2- 2^{10}*3^{11}*7^4\)?

A. 5
B. 9
C. 10
D. 12
E. 23


Here's my answer:

\(6^{12}*49^2- 2^{10}*3^{11}*7^4\)

\(6^{12}*49^2- 2^{10}*3^{10}*7^4*3^{1}\) ... where \(2^{10}*3^{10}\) = \(6^{10}\)

Common factor:

\(6^{10}*7^4((6^{2}*1) - (1*1*3^{1}))\)

Solving:

\(6^{10}*7^4(36 - 3)\)

\(6^{10}*7^4*(33)\)... where \(33 = 11^{1}*3^{1}\)

\(6^{10}*7^{4}*11^{1}*3^{1}\) ... where \(2^{10}*3^{10}\) = \(6^{10}\)

\(2^{10}*3^{11}*7^{4}*11^{1}\) ... where \(3^{10}*3^{1}\) = \(3^{11}\)

So at the end just sum:

2 + 3 +7 + 11 = 23

E

I know its a bit large but it is because I wanted to point out every step, when I was solving this type of questions for the first time it was difficult for me so I want to help new people :).
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What is the sum of all distinct prime factors of 6^12*49^2- 2^10*3^11*   [#permalink] 13 Feb 2019, 18:24
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