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Aside: Chetan noticed that, since the product ab has an ODD number of factors, we can conclude that ab is a perfect square.
For more on this concept, see our free video: https://www.gmatprepnow.com/module/gmat- ... /video/829
Here's a practice question to reinforce your learning: https://www.gmatprepnow.com/module/gmat- ... /video/830

Cheers,
Brent
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Hei folks,
I don't get where it said "the product is less than 100".
The numbers being multiplied are specified as <=100.

Cmon guys,
where did we get that from?
chetan2u and EMPOWERgmatRichC please throw light.



Hei folks,
I don't get where it said "the product is less than 100".
The numbers being multiplied are specified as <=100.

Cmon guys,
where did we get that from?
chetan2u and EMPOWERgmatRichC please throw light.



Hei folks,
I don't get where it said "the product is less than 100".
The numbers being multiplied are specified as <=100.

Cmon guys,
where did we get that from?
chetan2u and EMPOWERgmatRichC please throw light.



Hei folks,
I don't get where it said "the product is less than 100".
The numbers being multiplied are specified as <=100.

Cmon guys,
where did we get that from?
chetan2u and EMPOWERgmatRichC please throw light.
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Nez
Hei folks,
I don't get where it said "the product is less than 100".
The numbers being multiplied are specified as <=100.

Cmon guys,
where did we get that from?
chetan2u and EMPOWERgmatRichC please throw light.

Hi Nez,

the Q is
Quote:
If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100} and multiplied, what is the probability that the resulting number has EXACTLY 3 factors?

A. 4/(25*99)
B. 2/(25*99)
C. 8/(25*99)
D. 16/(25*99)
E. 32/(25*99)

why should be the product be <100..
the condition the resulting number has EXACTLY 3 factors? narrows down the answer to perfect square of a prime number..
so factor of that number= 1*p*p, where p is the prime number..
the other condition is a and b are different number, so 1*p*p can have two different factors as 1*p^2, where a=1 and b=p^2 or vice versa...
p*p would mean both a and b are same, which means a=b=p.. BUT this is not true as per the restrictions..
that is why a*b will have one of a or b as 1, and the other number perfect square <100.. so its product will always be <100..
hope it helps
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chetan2u
Bunuel
If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100} and multiplied, what is the probability that the resulting number has EXACTLY 3 factors?

A. 4/(25*99)
B. 2/(25*99)
C. 8/(25*99)
D. 16/(25*99)
E. 32/(25*99)

hi,
bunuel a good Q..

what does three factor means?
it means the number is a square of a prime number.

THE CATCH IS :- How do we get two different integers a and b such that the multiple of these numbers is a perfect square?

one number has to be 1 and other the perfect square itself..

lets see how many perfect square of which all prime numbers are <100...
2,3,5,7.. next 11 will lead to 121, which is >100..
if say a is one in every case b will be 4,9,25,49.... so 4 ways
vice versa.. b is 1 and a is 4,9,25,49..
so total ways =2*4=8...

total ways = 100*99..

prob= \(\frac{8}{{100*99}}\)..
prob= \(\frac{2}{{25*99}}\)
B

You list the total # of ways of selecting 2 numbers as 100*99, but isn't that incorrect?

100C2 = 50*99

Also, video solving the problem for those who need it.
https://gmatpractice.q-51.com/arithmetic ... ty-4.shtml
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chetan2u
Bunuel
If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100} and multiplied, what is the probability that the resulting number has EXACTLY 3 factors?

A. 4/(25*99)
B. 2/(25*99)
C. 8/(25*99)
D. 16/(25*99)
E. 32/(25*99)

hi,
bunuel a good Q..

what does three factor means?
it means the number is a square of a prime number.

THE CATCH IS :- How do we get two different integers a and b such that the multiple of these numbers is a perfect square?

one number has to be 1 and other the perfect square itself..

lets see how many perfect square of which all prime numbers are <100...
2,3,5,7.. next 11 will lead to 121, which is >100..
if say a is one in every case b will be 4,9,25,49.... so 4 ways
vice versa.. b is 1 and a is 4,9,25,49..
so total ways =2*4=8...

total ways = 100*99..

prob= \(\frac{8}{{100*99}}\)..
prob= \(\frac{2}{{25*99}}\)
B

You list the total # of ways of selecting 2 numbers as 100*99, but isn't that incorrect?

100C2 = 50*99

Also, video solving the problem for those who need it.
https://gmatpractice.q-51.com/arithmetic ... ty-4.shtml

HI,
100C2 means combination of two numbers in 100 numbers..
But here you have a then b and b, then a are two different selections..
first number can be selected in 100 ways..
when the first one is selected, the remaining are 99, out of which you have to select the 2nd one..
so ways = 100*99
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Guys, can anyone explain me why numbers such as 22 (1*2*11) or 51 (3*17*1) are not in your list of possible outcomes? If we make a matrix of prime numbers from 2 to 47 both in horizontal and vertical we'll find out 31 numbers between 1 and 100 that would fit this rule of having 3 factors and with products below 100:

2*2, 2*3, 2*5, 2*7, 2*11.... 2*47
3*3, 3*5, 3*7,.... 3*31
5*5, 5*7.... 5*17
7*7, 7*11, 7*13

Tks a lot in advance.
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livio04
Guys, can anyone explain me why numbers such as 22 (1*2*11) or 51 (3*17*1) are not in your list of possible outcomes? If we make a matrix of prime numbers from 2 to 47 both in horizontal and vertical we'll find out 31 numbers between 1 and 100 that would fit this rule of having 3 factors and with products below 100:

2*2, 2*3, 2*5, 2*7, 2*11.... 2*47
3*3, 3*5, 3*7,.... 3*31
5*5, 5*7.... 5*17
7*7, 7*11, 7*13

Tks a lot in advance.

HI,
why we do not take numbers like 22 or 51 is bcause they have more than 3 factors..
22- 1,2,11,22- 4 factors
51-1,3,17,52- 4 factors..
as explained above only square of prime number fit into this category..
hope it helped
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if a number has exactly 3 factors means, it is a square of a prime number.
take example prime 2 square number is 4. factors are 1 ,2 and4.
same for 9 is 1,3 and 9.

among numbers from 1to 100. we are selecting two distinct integers a and b and multiply it.

so two distinct integers with 3 factors one number must be 1 and other numbers are 4,9,25, and 49.

so total 4 pairs of combination available.
number of selctions of 2 numbers from 1 to 100 is 100c2

so probability= favourable events/number of events

=4/100*99/2=4/50*99
=2/25*99. so option B is correct.
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livio04
Guys, can anyone explain me why numbers such as 22 (1*2*11) or 51 (3*17*1) are not in your list of possible outcomes? If we make a matrix of prime numbers from 2 to 47 both in horizontal and vertical we'll find out 31 numbers between 1 and 100 that would fit this rule of having 3 factors and with products below 100:

2*2, 2*3, 2*5, 2*7, 2*11.... 2*47
3*3, 3*5, 3*7,.... 3*31
5*5, 5*7.... 5*17
7*7, 7*11, 7*13

Tks a lot in advance.

HI,
why we do not take numbers like 22 or 51 is bcause they have more than 3 factors..
22- 1,2,11,22- 4 factors
51-1,3,17,52- 4 factors..
as explained above only square of prime number fit into this category..
hope it helped

Thank you! It helped indeed! :wink:
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Bunuel
If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100} and multiplied, what is the probability that the resulting number has EXACTLY 3 factors?

A. 4/(25*99)
B. 2/(25*99)
C. 8/(25*99)
D. 16/(25*99)
E. 32/(25*99)


Lets consider 121 (perfect square of 11) -> 3 factors of 121 = 1, 11, 121 (which would be the possible values of a and b)

So a and b = (121 and 1) or (11 and 11) - but a, b <100 and also distinct. Hence the perfect squares need to be <100.

Hope this helps.
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Hello,

following up on the answers above, once you realize that one of the numbers has to be 1 and the others have to be squares of prime numbers (in this case, 4, 9, 25, 49), you can use the hypergeometric distribution as a quick way to figure out the probability:

In this case, you're finding the probability of choosing 1 from 1 number (1 - 1C1) AND 1 from 4 numbers (4,9,25,49 - 4C1), from a total 2 out of a total of 100 numbers (100C2).

numerator: The possible number of event outcomes - in this case, the number of outcomes will be 1C1 * 4C1 (just 4)
denominator: The total possible outcomes - 100C2 ((99 * 100) / 2 or 99*50)

So the total probability would be (1C1 * 4C1) / (100C2) = 4 / (99 * 50) = = 2 / (99 * 25)

hope this helps
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Bunuel
If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100} and multiplied, what is the probability that the resulting number has EXACTLY 3 factors?

A. 4/(25*99)
B. 2/(25*99)
C. 8/(25*99)
D. 16/(25*99)
E. 32/(25*99)

I feel here that the resultant of multiplication is important rather than the order.
Total ways of selecting number with EXACTLY 3 factors (1,4) (1,9) (1,25) (1,49) = 4
Total ways of choosing the number = \(100C2\)= \(50*99\)
Hence, Probability = \(4/50*99\) = \(2/25*99\)
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Bunuel,
I did get the answer right but it was more of a guess than actual solving, if possible could you share the method you would have used to solve the question ?
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I think here arrangement of numbers as u said " a then b and b, then a " doesn't matter but only the selection ,so we can take total selection as 10C2 and no of fav outcome =4,thus prob= 4/10C2
Plz confirm if i am correct or not
chetan2u


HI,
100C2 means combination of two numbers in 100 numbers..
But here you have a then b and b, then a are two different selections..
first number can be selected in 100 ways..
when the first one is selected, the remaining are 99, out of which you have to select the 2nd one..
so ways = 100*99
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