If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100

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

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.

Hi Nez,

the Q is


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

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

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

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.

Thank you! It helped indeed!

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.

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

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 = \(4/50*99\) = \(2/25*99\)

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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