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

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.

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