Question

Two numbers p and q are chosen at random from the set of first 30 natural numbers. What is the probability that p² - q² is divisible by 3?

Answer 1

Answer:

18/29

Step-by-step explanation:

Hi,

Given that two numbers p and q are chosen from the set of first

30 natural numbers , possible number of ways in which these 2

could be selected are ³⁰C₂ , but p and q can get interchanges

among themselves in 2 different ways . Hence, total number of

ways are 2*³⁰C₂ = 870

If (p² - q²) has to be divisible by 3 ,

Since p² - q² = (p - q)*(p + q)

Either p - q should be a multiple of 3 or p + q should be a

multiple of 3

Case 1: If both p and q are multiples of 3

Since, there are 10 multiples of 3 among the first 30 natural

numbers, p and q can be selected in ¹⁰C₂ ways and p and q can

interchange among themselves in 2 different ways, hence total

number of selections are  2*¹⁰C₂ = 90.

Case 2: If both p and q are of the form 3k + 1

Since, there are 10 numbers  of the form 3k + 1 among the first

30 natural numbers, p and q can be selected in ¹⁰C₂ ways and p

and q can interchange among themselves in 2 different ways,

hence total number of selections are  2*¹⁰C₂ = 90 , If selection is

made in this way then difference of the numbers will be divisible

by 3 and hence the difference of their squares.

Case 3: If both p and q are of the form 3k + 2

Since, there are 10 numbers  of the form 3k + 2 among the first

30 natural numbers, p and q can be selected in ¹⁰C₂ ways and p

and q can interchange among themselves in 2 different ways,

hence total number of selections are  2*¹⁰C₂ = 90.

If selection is made in this way then difference of the numbers

will be divisible by 3 and hence the difference of their squares.

Case 4: Either one of p and q is of the form 3k + 1 and the other

is of the form 3k + 2. Since, there are 10 numbers of the form

3k + 1 and 10 numbers of the form 3k + 2 among the first 30

natural numbers, one among p and q can be selected in ¹⁰C₁

ways (the number of form 3k + 1 and  the other can be

selected  in ¹⁰C₁ ways (the number of form 3k + 2) and p and q

can interchange among themselves in 2 different ways, hence

total number of selections are  2*¹⁰C₁*¹⁰C₁ = 200.

If selection is made in this way then sum of the numbers will be

divisible by 3 and so is  the difference of their squares.

Total number of ways of selecting p and q such that p² - q² is

divisible by 3 are

= ( 90 + 90 + 90 + 270 )

= 540

So, the probability that  p² - q² is divisible by 3 will be

= 540/870

= 18/29

Hope, it helps !

Answer 2

Answer:

47/87

Step-by-step explanation:

Note that a2−b2 is divisible by 3 if and only if either (i) a and b are both divisible by 3 or (ii) neither a nor b is divisible by 3. This is because if n is not divisible by 3, then n has remainder 1 or 2 on division by 3. If a and b have the same remainder on division by 3, then 3 divides a−b. And if one has remainder 1 and the other 2, then 3 divides a+b. Finally, a2−b2=(a−b)(a+b).

There are (10C2) choices of Type (i), and (20C2) of Type (ii).

Alternately, we count the bad pairs, where one of the numbers is divisible by 3 and the other is not. There are (10)(20) bad pairs.