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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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Math Expert
Joined: 02 Sep 2009
Posts: 53020
What is the sum of all distinct prime factors of 6^12*49^2- 2^10*3^11*  [#permalink]

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28 Jan 2019, 02:31
00:00

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 $$6^{12}*49^2- 2^{10}*3^{11}*7^4$$?

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

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Joined: 09 Mar 2018
Posts: 1001
Location: India
What is the sum of all distinct prime factors of 6^12*49^2- 2^10*3^11*  [#permalink]

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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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Joined: 13 Jan 2018
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Location: India
Concentration: Operations, General Management
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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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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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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.

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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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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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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

$$6^{12}*49^2- 2^{10}*3^{11}*7^4$$

$$6^{12}*49^2- 2^{10}*3^{10}*7^4*3^{1}$$ ... 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 .
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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