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Jack has a few six­faced dice. Each die has numbers 3, 5, and 6 each

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Jack has a few six­faced dice. Each die has numbers 3, 5, and 6 each [#permalink]

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Jack has a few six­faced dice. Each die has numbers 3, 5, and 6 each printed on two faces. He throws all dice and multiplies
the digits displayed on the top surface of each die. If the product he got is 243,000, how many dice does Jack have?

A. 8
B. 11
C. 12
D. 14
E. More than 14

Source: www.expertsglobal.com
[Reveal] Spoiler: OA

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Re: Jack has a few six­faced dice. Each die has numbers 3, 5, and 6 each [#permalink]

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New post 11 Oct 2017, 02:15
243000 = 2^3 * 3^5 * 5^3, which can be re-written as (2*3)^3 * 3^2 * 5^3 or 6^3 * 3^2 * 5^3
So total dice throw required as a minimum is 3+2+3 = 8.
Option A

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Re: Jack has a few six­faced dice. Each die has numbers 3, 5, and 6 each [#permalink]

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New post 11 Oct 2017, 02:23
nkmungila wrote:
Jack has a few six­faced dice. Each die has numbers 3, 5, and 6 each printed on two faces. He throws all dice and multiplies
the digits displayed on the top surface of each die. If the product he got is 243,000, how many dice does Jack have?

A. 8
B. 11
C. 12
D. 14
E. More than 14

Source: http://www.expertsglobal.com


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Hope it helps.
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Re: Jack has a few six­faced dice. Each die has numbers 3, 5, and 6 each [#permalink]

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nkmungila wrote:
Jack has a few six­faced dice. Each die has numbers 3, 5, and 6 each printed on two faces. He throws all dice and multiplies
the digits displayed on the top surface of each die. If the product he got is 243,000, how many dice does Jack have?

A. 8
B. 11
C. 12
D. 14
E. More than 14

Source: http://www.expertsglobal.com


You are given that in each die, two faces have 3, two faces have 5 and two faces have 6.

243000 = 2^3 * 3^5 * 5^3

But note that we have multiplied only 3s, 5s and 6s together. So all 2s must be clubbed with 3s to get 6s.

So we have three 6s, two leftover 3s and three 5s.

This gives us a total of 3 + 2 + 3 = 8 dice

Answer (A)
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Re: Jack has a few six­faced dice. Each die has numbers 3, 5, and 6 each [#permalink]

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New post 13 Oct 2017, 08:22
nkmungila wrote:
Jack has a few six­faced dice. Each die has numbers 3, 5, and 6 each printed on two faces. He throws all dice and multiplies
the digits displayed on the top surface of each die. If the product he got is 243,000, how many dice does Jack have?

A. 8
B. 11
C. 12
D. 14
E. More than 14


et’s break 243,000 into prime factors:

243,000 = 243 x 1000 = 3^5 x 10^3 = 3^5 x 2^3 x 5^3

We have the prime factorization of 243,000; however, since there are only 3s, 5s, and 6s (but not 2s) on the faces of the dice, the factor 2 must be a factor of 6. Thus, we have to redistribute the factors so that the factorization only contains factors of 3, 5, and 6:

243,000 = 3^5 x 2^3 x 5^3 = 3^2 x 3^3 x 2^3 x 5^3 = 3^2 x 6^3 x 5^3

Since now all the factors are 3s, 5s, and 6s, the sum of the exponents will be the number of dice:

2 + 3 + 3 = 8

Therefore, there are 8 dice.

Note: If you are uncertain about why the number of dice is equal to the sum of the exponents, think of more straightforward outcomes of dice rolls: If the product of the dice rolls were, say, 125, you would know that each of the 3 dice showed a 5. Expressing 125 as 5^3, you see that the exponent 3 tells how many dice were rolled. Similarly, if the product of the dice rolls were, say, 135, then you would factor 135 as 5 x 27 = 5^1 x 3^3, and this would indicate that there were 1 + 3 = 4 dice and they showed 5,3,3,3 on their faces.

Answer: A
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Re: Jack has a few six­faced dice. Each die has numbers 3, 5, and 6 each   [#permalink] 13 Oct 2017, 08:22
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