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Each factor of 210 is inscribed on its own plastic ball, and [#permalink]

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30 Apr 2012, 07:22

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Each factor of 210 is inscribed on its own plastic ball, and all of the balls are placed in a jar. If a ball is randomly selected from the jar, what is the probability that the ball is inscribed with a multiple of 42?

Each factor of 210 is inscribed on its own plastic ball, and all of the balls are placed in a jar. If a ball is randomly selected from the jar, what is the probability that the ball is inscribed with a multiple of 42?

A. 1/16 B. 5/42 C. 1/8 D. 3/16 E. 1/4

Please post the fastest method with time.

210=2*3*5*7, so the # of factors 210 has is (1+1)(1+1)(1+1)(1+1)=16 (see below); 42=2*3*7, so out of 16 factors only two are multiples of 42: 42 and 210, itself;

So, the probability is 2/16=1/8.

Answer: C.

Finding the Number of Factors of an Integer

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.

Re: Each factor of 210 is inscribed on its own plastic ball, and [#permalink]

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30 May 2012, 22:23

maffo wrote:

Sorry Bunuel,

i don't get it.

as you said:

210 = 2*5*7*3 and 42 = 2*7*3; so i would say there are 5 multiples of 42 between 42 and 210.

since I'm obviously wrong, can you please help me to understand where's the mistake?

thanks in advance

By the time we are waiting for Bunuel's reply , I will attempt to answer this. I think you mean 42*1, 42*2 ... 42*5 as 5 multiples. But the plastic balls that we have with us, do not have these numbers inscribed on them, because they are not the factors of 210.

For 42*2 to be on one of the plastic balls, we will need an additional 2 with 42 (which is not there, we only have a 5 left; 42 already takes care of 2*7*3 from 210.)

So, 42*2 = 84 , is not a factor of 210, therefore the assumption does not hold true.

210 = 2*5*7*3 and 42 = 2*7*3; so i would say there are 5 multiples of 42 between 42 and 210.

since I'm obviously wrong, can you please help me to understand where's the mistake?

thanks in advance

We are not interested in the multiples of 42 between 42 and 210. We are interested in multiples of 42 which are factors of 210. Out of factors of 210 only 2 are multiples of 42: 42 and 210, itself.
_________________

Re: Each factor of 210 is inscribed on its own plastic ball, and [#permalink]

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31 May 2012, 04:05

Bunuel wrote:

BDSunDevil wrote:

Each factor of 210 is inscribed on its own plastic ball, and all of the balls are placed in a jar. If a ball is randomly selected from the jar, what is the probability that the ball is inscribed with a multiple of 42?

A. 1/16 B. 5/42 C. 1/8 D. 3/16 E. 1/4

Please post the fastest method with time.

210=2*3*5*7, so the # of factors 210 has is (1+1)(1+1)(1+1)(1+1)=16 (see below); 42=2*3*7, so out of 16 factors only two are multiples of 42: 42 and 210, itself;

So, the probability is 2/16=1/8.

Answer: C.

Finding the Number of Factors of an Integer

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.

Each factor of 210 is inscribed on its own plastic ball, and all of the balls are placed in a jar. If a ball is randomly selected from the jar, what is the probability that the ball is inscribed with a multiple of 42?

A. 1/16 B. 5/42 C. 1/8 D. 3/16 E. 1/4

Please post the fastest method with time.

210=2*3*5*7, so the # of factors 210 has is (1+1)(1+1)(1+1)(1+1)=16 (see below); 42=2*3*7, so out of 16 factors only two are multiples of 42: 42 and 210, itself;

So, the probability is 2/16=1/8.

Answer: C.

Finding the Number of Factors of an Integer

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.

Re: Each factor of 210 is inscribed on its own plastic ball, and [#permalink]

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21 Jul 2014, 17:41

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Re: Each factor of 210 is inscribed on its own plastic ball, and [#permalink]

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01 Apr 2015, 19:03

Finding the number of factors is the same as put out by Bunuel in the example.

Now to find the multiple of 42 in 210 we can do the following: write the factors as 2x3x5x7 as (2^0 + 2^1)(3^0 + 3^1)(5^0 + 5^1)(7^0 + 7^1) {this is the way of finding the sum of the number of factors of a number) Now look carefully and see the number combination that brings in 42 into the question. remove (2^0), (3^0) and (7^0) so that we are left with (2^1)(3^1)(7^1)(5^0+5^1) ie 42(5^0+5^1). The above means that there are only two such numbers. Hence the probability will be 2/16 = 1/8

Re: Each factor of 210 is inscribed on its own plastic ball, and [#permalink]

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22 Jun 2015, 22:35

1

This post received KUDOS

Please tag Probability

BDSunDevil wrote:

Each factor of 210 is inscribed on its own plastic ball, and all of the balls are placed in a jar. If a ball is randomly selected from the jar, what is the probability that the ball is inscribed with a multiple of 42?

Each factor of 210 is inscribed on its own plastic ball, and [#permalink]

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25 Feb 2016, 22:04

Hi, I had the same question as GMATDeep the second time around reviewing this question. Perhaps this is a slightly different way to think about it...

210=2*3*5*7, so the # of factors 210 has is (1+1)(1+1)(1+1)(1+1)=16 (see Bunuel's formula); 42=2*3*7, so out of 16 factors are: (2*3*7)*5^0 = 42 and (2*3*7)*5^1 = 210;

Re: Each factor of 210 is inscribed on its own plastic ball, and [#permalink]

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18 Sep 2016, 18:55

Bunuel wrote:

BDSunDevil wrote:

Each factor of 210 is inscribed on its own plastic ball, and all of the balls are placed in a jar. If a ball is randomly selected from the jar, what is the probability that the ball is inscribed with a multiple of 42?

A. 1/16 B. 5/42 C. 1/8 D. 3/16 E. 1/4

Please post the fastest method with time.

210=2*3*5*7, so the # of factors 210 has is (1+1)(1+1)(1+1)(1+1)=16 (see below); 42=2*3*7, so out of 16 factors only two are multiples of 42: 42 and 210, itself;

So, the probability is 2/16=1/8.

Answer: C.

Finding the Number of Factors of an Integer

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.

Each factor of 210 is inscribed on its own plastic ball, and all of the balls are placed in a jar. If a ball is randomly selected from the jar, what is the probability that the ball is inscribed with a multiple of 42?

A. 1/16 B. 5/42 C. 1/8 D. 3/16 E. 1/4

Please post the fastest method with time.

210=2*3*5*7, so the # of factors 210 has is (1+1)(1+1)(1+1)(1+1)=16 (see below); 42=2*3*7, so out of 16 factors only two are multiples of 42: 42 and 210, itself;

So, the probability is 2/16=1/8.

Answer: C.

Finding the Number of Factors of an Integer

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.

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