IIT JEE , AIEEE Forum - Mathematics , Algebra - IIT 2006 question... If r,s,t are prime numbers and p,q are +ve integers sch that lcm of

vaibhav_manu

IIT 2006 question... If r,s,t are prime numbers and p,q are +ve integers sch that lcm of p,q is r²s⁴t²... then the number of ordered pairs p,q ar...

plzz help m findin disparity wid the ans...

konichiwa2x

Nice question...No wonder!

x must be of the form

, where

and

.

Similarly

and

.

Furthermore, at least one of

,

and

must be equal to 5, and at least one of

,

and

must be equal to 3.

Look at the a's first. Either one of them, or two of them, or all three of them, must be equal to 5.

In the first case (when just one of them is equal to 5), there are three ways of choosing which one is equal to 5, and for each of these there are

ways of assigning a value

to the other two a's.

If two of the a's are equal to 5, then again there are three ways in which this can happen, and for each of these there are 5 possible values for the "a" that is not equal to 5.

Finally, if all three a's are equal to 5, there is only one way of doing this. Thus the total number of possible ways of assigning values to

,

and

is

Now look at the possible values for

,

and

. We can use the same analysis as before, except that this time one or more of the b's must be equal to 3, and the remaining b's (if any) have to take one of the three values 0, 1 or 2. Thus the total number of possible ways of assigning values to

,

and

is

So, putting the a's and b's together, the answer to the problem should be that there are

possible triples

.

Edit: The answer is gave above is when

are distinct primes and you restrict

to be positive integers. Multiply that by

if you want to drop the positivity assumption. If you want to drop the condition

distinct prime you will need to give the prime factorisation of r and t.

yuiop

hey, is the answer 360?

hsbhatt

Even i am getting 360

Basically, the numbers r², s⁴ and t² have to appear in either number. This can be assigned in 2*2*2 = 8 ways.

Now if r² appears in one number, the other can have r⁰or r¹ or r² as its factors.

This reasoning gives 8*3*5*3 = 360 such ordered pairs

elastiboysai

Hi guyz

I too followed a similiar analogy but

I am gettin a different ans- 225

Cases :numbers denote exponent

for r in p q

r 0 2

1 2

2 0,1,2

total 5 cases

for t in p q same way 5 cases

and for s in p q

s 0 4

1 4

2 4

3 4

4 0,1,2,3,4

total 9 cases

So total cases= 5*9*5= 225

jaggy

ans is 225.ihave solved it earlier

hsbhatt

I can see my mistake now when you enumerate it. I was counting the case when both p and q have have r² as a factor twice. So that's alright then.

sboosy

elastic and jagery are exactly correct the answer is 225

amitp91

ANSWER IS 225

vaibhav_manu

i think the ans is 223... cuz the pair (1,1) is being repeated thrice... plzz chk...

cursed_mask

yeah the answer is 225...this question is a particular case of this general question i came across yesterday

http://www.locuseducation.org/component/option,com_fireboard/Itemid,26/func,view/catid,5/id,280/#280

vaibhav_manu

plzz check on what im finding out...

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