**#Problem 13.1**

Let n ≥ 1 be an integer and let t denote the number of positive integer divisors of n^2. Show

that the number of positive integer solutions (a, b) of the equation 1/a−1/b = 1/n is precisely

equal to (t − 1)/2.

that the number of positive integer solutions (a, b) of the equation 1/a−1/b = 1/n is precisely

equal to (t − 1)/2.

**Solution scheme and approach:**

1/a-1/b=1/n..............(1)

From the above equation it's clear that 1/a>1/n =>n>a

Now we can consider an integer x such that

a=n-x

From equation number (1) we can also write

1/b=1/(n-x)-1/n=x/(n-x)*n

=>b=(n^2-nx)/x=n^2/x-n

To count the total number of solutions we have to count total numbers of possible values of x.

As b is an integer x has to be an integer which must be a factor of n^2 such that x

Now think, n^2 is a perfect square which has a total of t divisors. out of these factors one is n another (t-1)/2 is less than n each of them is paired with other (t-1)/2 number of factors which are greater than n.

[Suppose for 16=4x4=2x8=1x16]

**Hence, Total number of solutions are (t-1)/2. Q.E.D.**

[N.B: Total number of ordered pair(a,b) would be (t-1)]

Food for thought:

Let n ≥ 1 be an integer and let t denote the number of positive integer divisors of n^2. Show that the number of positive integer solutions (a, b) of the equation 1/a+1/b = 1/n is precisely equal to t /2.

**#Problem 13.2**

Find the smallest positive integer n with the property that the equation 1/a − 1/b = 1/n has exactly 2010 different solutions in positive integers a and b.

(a)6^2010,(b)2^2010,(c)1,(d)10^2010,(e)Cannot be determined

**Solution scheme and approach**

From the above equation it's clear that

n^2 should have t number of divisors where (t-1)/2=2010=>t=4021

Suppose

n=p1^a*p2^b*p3^c*...

=>n^2=p1^2a*p2^2b*p3^2c*...pn^2n

It has (2a+1)*(2b+1)*(2c+1)*....*(2n+1)=4021 number of divisors

Now the catch is 4021 is a prime number.=>n^2=p1^2a

=>(2a+1)=4021

=>a=2010

As we need to find out smallest possible value of p1 that means p1=2

So the required number is 2^2010

**Hence, option (b) is the right option.**

**#Problem 13.3**

(a)14, (b)8, (c)15, (d)7,(e)16 [Career Launcher Mock CAT]

**Solution scheme and approach**

Ramu and Krishna work 1/R and 1/K th work per day respectively

According to the question as they are working on alternative day.We may say if they work together work must be completed on 12th day

=>12(1/R+1/K)=1

=>(1/R+1.K)=1/12

As, 12^2=2^4*3^2 has [(4+1)*(2+1)]/2=15/2 number of solutions.So total number of ordered pair(R,K) solutions are 15. (refer Food for thought)

Hence, Option (c).

Hope, this session has helped you a little bit to clear these things.

All the best!!

awesome concept thanks :)

ReplyDeleteand in the last question

t is 15 so t/2 , the no of pairs of (R,K) should be 7 right

No of pairs are (R,k) is 7 no doubt for these are not ordered.

ReplyDeleteAs u know (R,K) can be also (K,R) so total ordered pairs are 15/2*2=15.

If still it's not clear let me know.

atb!!

got it :) thanks

ReplyDelete