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

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

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.

Hope I am making some sense !!
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Re: Each factor of 210 is inscribed on its own plastic ball, and [#permalink]
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.

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

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.

For more on these issues check Number Theory chapter of Math Book: math-number-theory-88376.html

Hope it helps.

As it said multiples of 42, I had not considered 42 itself.
Where did I went wrong in reasoning?
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Re: Each factor of 210 is inscribed on its own plastic ball, and [#permalink]
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manulath wrote:
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.

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.

For more on these issues check Number Theory chapter of Math Book: math-number-theory-88376.html

Hope it helps.

As it said multiples of 42, I had not considered 42 itself.
Where did I went wrong in reasoning?

Note that an integer $$a$$ is a multiple of an integer $$b$$ (integer $$a$$ is a divisible by an integer $$b$$) means that $$\frac{a}{b}=integer$$.

Since 42/42=integer then 42 is a multiple of itself (generally every positive integer is a multiple of itself).

Hope it's clear.
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Re: Each factor of 210 is inscribed on its own plastic ball, and [#permalink]
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
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Re: Each factor of 210 is inscribed on its own plastic ball, and [#permalink]
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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.
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Each factor of 210 is inscribed on its own plastic ball, and [#permalink]
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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;

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

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

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.

For more on these issues check Number Theory chapter of Math Book: math-number-theory-88376.html

Hope it helps.

Why are we not considering negative values?
-42 is also a multiple ?!
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Re: Each factor of 210 is inscribed on its own plastic ball, and [#permalink]
sidoknowia wrote:
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.

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.

For more on these issues check Number Theory chapter of Math Book: math-number-theory-88376.html

Hope it helps.

Why are we not considering negative values?
-42 is also a multiple ?!

Factor is a positive divisor.
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Re: Each factor of 210 is inscribed on its own plastic ball, and [#permalink]
Got to read the question carefully, I read it as "what is the probability that the ball is inscribed with 42?"

And calculated option A as my answer
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Re: Each factor of 210 is inscribed on its own plastic ball, and [#permalink]
Hi All,

We're told that each factor of 210 is inscribed on its own plastic ball, and all of the balls are placed in a jar and a ball is randomly selected from the jar. We're asked for the probability that the ball is inscribed with a multiple of 42. This question can be solved with some basic Arithmetic and 'brute force.'

Determining the factors of 210 isn't too difficult (and there are some great 'math shortcuts' that can save you some time if you know the rules of division). The factors are:
1 and 210
2 and 105
3 and 70
5 and 42
6 and 35
7 and 30
10 and 21
14 and 15

Thus, there are 16 total factors. Looking at this list, the only ones that are MULTIPLES of 42 are 42 and 210... so 2 of the 16 = 2/16 = 1/8

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

Let’s break 210 into primes:

210 = 21 x 10 = 3 x 7 x 5 x 2

To find the total number of factors, we add 1 to each exponent of the prime factors and then multiply those values. Since each prime factor of 210 has an exponent of 1, we have:
(1 + 1)(1 + 1)(1 + 1)(1 + 1) = 16 total factors

There are 2 multiples of 42, and they are:

2 x 3 x 7

2 x 3 x 7 x 5

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

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Each factor of 210 is inscribed on its own plastic ball, and [#permalink]
Bunuel ,
1. I LOVED your formula, tnx!
2. If I didn't know it, could I use a probability approach?

So I though this way:

The first factor could be ether of those 4:
1
2
3
7
But it can't be:
5

So we have:

__ __ __ __
$$\frac{4}{5}$$*$$\frac{3}{4}$$*$$\frac{2}{3}$$*$$\frac{1}{2}$$=$$\frac{1}{5}$$

Is it possible to sove like this? If so, where did I make a mistake?
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Re: Each factor of 210 is inscribed on its own plastic ball, and [#permalink]
210=2.3.5.7

so number of factors = 2.2.2.2=16

multiples of 42 = 42,210 (2 nos)

therefore probability = 2/16=1/8
Re: Each factor of 210 is inscribed on its own plastic ball, and [#permalink]
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