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In a certain game, a large container is filled with red, yel [#permalink]

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29 Dec 2012, 06:10

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In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed?

In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed?

(A) 5 (B) 4 (C) 3 (D) 2 (E) 0

\(147,000=2^3*3*5^3*7^2\).

Since no other beads value is a multiple of 7, then there must be two red beads removed.

Re: In a certain game, a large container is filled with red, yel [#permalink]

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08 Feb 2013, 11:31

It is easy after you recognize that you need to do a prime factorization. To make it easier the point values are prime numbers too. Tricky but we can do it!
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Re: In a certain game, a large container is filled with red, yel [#permalink]

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21 Feb 2014, 12:28

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In a certain game, a large container is filled with red, yel [#permalink]

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22 May 2014, 11:18

In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed? A. 5 B. 4 C. 3 D. 2 E. 0

I am not able to comprehend the question correctly. There could be a lot of combinations for removing the beads. How could we find any particular answer?

In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed? A. 5 B. 4 C. 3 D. 2 E. 0

I am not able to comprehend the question correctly. There could be a lot of combinations for removing the beads. How could we find any particular answer?

Merging similar topics. please refer to the discussion above and ask if anything remains unclear.

Re: In a certain game, a large container is filled with red, yel [#permalink]

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22 May 2014, 12:15

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Dear amarmeena1992

This is not a Combinations question. (And it cannot be, since you are not even told what the total number of balls is.)

You are told that the product of the scores of all the balls that were taken out is 147000. Now, \(147000 = 7^2*5^3*3*4*2^3\)

From this expression, you can easily deduce that:

2 Red balls were taken out (since each Red ball has a score of 7. So, number of Red balls = power of 7 in the expression), along with 3 Yellow balls, 1 Green ball and 3 Blue balls.

Hope this helps.
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Last edited by JapinderKaur on 22 May 2014, 14:37, edited 2 times in total.

Re: In a certain game, a large container is filled with red, yel [#permalink]

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22 May 2014, 13:01

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MavenQ wrote:

Dear amarmeena1992

This is not a Combinations question. (And it cannot be, since you are not even told what the total number of balls is.)

You are told that the product of the scores of all the balls that were taken out is 147000. Now, \(147000 = 7^2*5*3*4*2^3\)

From this expression, you can easily deduce that:

2 Red balls were taken out, along with 3 Yellow balls, 1 Green ball and 3 Blue balls.

Hope this helps.

Hi MavenQ, Kindly correct the factors : \(147,000 = 7^2*5^3*3*2^3\)

which would mean 2 R, 3 Y , 1 G and 3 B beads

Please press kudos if it helps you

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Dear amarmeena1992, The beads R,Y,G,B are allotted some numbers which are prime. So now since the product of the values of removed beads are known, we can decompose or factorize the product into unique combination/product of the bead values as shown : 147,000 = 7*7*3*5*5*5*2*2*2 = \(7^2*5^3*3*2^3\)

that is 2 R, 3 Y , 1 G and 3 B Kudos is the best form of appreciation

Re: In a certain game, a large container is filled with red, yel [#permalink]

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09 Jun 2014, 21:06

Bunuel wrote:

Walkabout wrote:

In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed?

(A) 5 (B) 4 (C) 3 (D) 2 (E) 0

\(147,000=2^3*3*5^3*7^2\).

Since no other beads value is a multiple of 7, then there must be two red beads removed.

Now I know the correct answer but can you help me to understand what conceptual mistake did I do initially.

In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed?

(A) 5 (B) 4 (C) 3 (D) 2 (E) 0

Initially I solved it like this:

14700 = 7 X 7 X 3 X 10^3

Since red bead is a multiple of 7 so I conclude that there were 7 red beads removed.
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In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed?

(A) 5 (B) 4 (C) 3 (D) 2 (E) 0

\(147,000=2^3*3*5^3*7^2\).

Since no other beads value is a multiple of 7, then there must be two red beads removed.

Now I know the correct answer but can you help me to understand what conceptual mistake did I do initially.

In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed?

(A) 5 (B) 4 (C) 3 (D) 2 (E) 0

Initially I solved it like this:

14700 = 7 X 7 X 3 X 10^3

Since red bead is a multiple of 7 so I conclude that there were 7 red beads removed.

Red beads worth 7 points. Say there were only 2 red beads removed. What would be the product of the point values of the removed beads? 7*7. If 7 red beads were removed the product would be 7*7*7*7*7*7*7.

In a certain game, a large container is filled with red, yel [#permalink]

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21 Jun 2015, 01:26

honchos wrote:

Bunuel wrote:

Walkabout wrote:

In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed?

(A) 5 (B) 4 (C) 3 (D) 2 (E) 0

\(147,000=2^3*3*5^3*7^2\).

Since no other beads value is a multiple of 7, then there must be two red beads removed.

Now I know the correct answer but can you help me to understand what conceptual mistake did I do initially.

In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed?

(A) 5 (B) 4 (C) 3 (D) 2 (E) 0

Initially I solved it like this:

14700 = 7 X 7 X 3 X 10^3

Since red bead is a multiple of 7 so I conclude that there were 7 red beads removed.

Initially I did the same mistake... But then I saw number 7 suits none of the answer choices but it should be the correct logic, as I did factorization, so it must be 2 One must read carefully, it says the PRODUCT of the removed beads, so 49 = 7*7 --> 2 Red beads
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In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed?

(A) 5 (B) 4 (C) 3 (D) 2 (E) 0

The first thing we want to do is to decipher the problem. The key word here is “product;" we are told that the product of the point values of the removed beads is 147,000.

Let's look at an example. Let’s say 2 red beads, 2 yellow beads, 2 green beads, and 2 blue beads were removed. The product of the point values would be:

7 x 7 x 5 x 5 x 3 x 3 x 2 x 2 = 44,100

Conversely, if we were to take 44,100 and break it down into its prime factors, we would get:

44,100 = 7 x 7 x 5 x 5 x 3 x 3 x 2 x 2

Note that this prime factorization of 44,100 corresponds exactly to the point values of the 2 red beads, the 2 yellow beads, the 2 green beads, and the 2 blue beads from the example.

This example is important because we can now use the same approach for answering the actual question. We are given that the product of the point values of the removed beads is 147,000. Thus, some number of 7’s, 5’s, 3’s, and 2’s is multiplied together to equal 147,000. Since we only care about the red beads, we only care about the number of 7’s in the product. Let’s keep this in mind as we break down 147,000.

147,000 = 147 x 1,000 = 49 x 3 x 1,000 = 7 x 7 x 3 x 1,000

Since 1,000 does not have 7 as a factor, we see that there are two 7’s in the product of 147,000; thus, 2 red beads were removed.

The answer is D.
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In a certain game, a large container is filled with red, yel [#permalink]

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05 May 2017, 13:47

I did the same thing. Prime Factorize 147 = 7^2 * 3. But then since to count number of factors we always add 1 so I did (2+1) = 3 as answer, instead of 2. I still dont understand why we dont add 1 to the power of 7.

Correct Answer is D, but my answer was C.
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