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# If p and q are two numbers and their LCM is (r square)(t square)(s raised to 4) where r, s, t are prime numbers. Then find the ordered pairs (p,q).Please Reply. AKASH GOYAL AskiitiansExpert-IITD
419 Points
10 years ago

Dear Arvind

p  and  q  are  two  positive  integers  whose  l.c.m.

is r2 s 4 t 2.  This ?rst of all means that neither p  nor q can have any prime

factor besides r, s and t.  So each of them is a product of powers of some

of these three primes.  We can therefore write p, q  in the form

p = ra sb tc    and    q = ru sv tw                             (1)

where a, b, c, u, v, w are non-negative integers.  Then the l.c.m., say e, of p and q is given by

e = ri  sj  tk                                                                                              (2)

where

i = max{a, u},       j = max{b, v}       and    k = max{c, w}                 (3)

This is the key idea of the problem.  The problem is now reduced to ?nding

the  number  of  triplets  of  ordered  pairs  of  the  form  {(a, u), (b, v), (c, w)}

where a, b, c, u, v, w are non-negative integers that satisfy

max{a, u} = 2,       max{b, v} = 4       and     max{c, w} = 2                (4)

Let us see in how many ways the ?rst entry of this triplet, viz., (a, u)

can be formed.  We want at least one of a and u to equal 2.  If we let a = 2,

then the possible values of u are 0, 1 and 2.  These are three possibilities.

Similarly,  with u  = 2  there will  be  three possibilities,  viz.        a = 0, 1 or 2.

So,  in  all  the  ?rst  ordered pair  (a, u)  can  be  formed  in  6  ways.      But  the

possibility (2, 2) has been counted twice.  So, the number of ordered pairs

of the type (a, u) that satisfy the ?rst requirement in (4) is 5 and not 6.

By  an  entirely  analogous  reasoning,  the  number  of  ordered  pairs  of

the  form  (b, v)  which  satisfy  the  second  requirement  in  (4)  is  2 × 5 − 1,

i.e.  9  while  that  of  ordered  pairs  of  the  type  (c, w)  satisfying  the  third

requirement in (4) is 5.  But the ways these three ordered pairs are formed

are completely independent of each other.  So the total number of triplets

of  ordered pairs  of the  form  {(a, u), (b, v), (c, w)} where  a, b, c, u, v, w  are

non-negative integers that satisfy (4) is 5 ×9 ×5 = 225.

All the best.

AKASH GOYAL

Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.

Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian. Kushagra Madhukar
one year ago
Dear student,

p  and  q  are  two  positive  integers  whose  LCM is r2 s 4 t 2
This first of all means that neither p  nor q can have any prime factor besides r, s and t.  So each of them is a product of powers of some of these three primes. We can therefore write p, q  in the form

p = ra sb tc    and    q = ru sv tw                             (1)

where a, b, c, u, v, w are non-negative integers.  Then the LCM, say e, of p and q is given by

e = ri  sj  tk                                                                                              (2)

where, i = max{a, u},       j = max{b, v}       and    k = max{c, w}                 (3)

This is the key idea of the problem.  The problem is now reduced to finding the  number  of  triplets  of  ordered  pairs  of  the  form  {(a, u), (b, v), (c, w)} where a, b, c, u, v, w are non-negative integers that satisfy

max{a, u} = 2,       max{b, v} = 4       and     max{c, w} = 2                (4)

Let us see in how many ways the ?rst entry of this triplet, viz., (a, u) can be formed.  We want at least one of a and u to equal 2.  If we let a = 2,
then the possible values of u are 0, 1 and 2.  These are three possibilities.
Similarly,  with u  = 2  there will  be  three possibilities,  viz.        a = 0, 1 or 2.
So,  in  all  the  first  ordered pair  (a, u)  can  be  formed  in  6  ways.
But  the possibility (2, 2) has been counted twice. So, the number of ordered pairs of the type (a, u) that satisfy the first requirement in (4) is 5 and not 6.
By  an  entirely  analogous  reasoning,  the  number  of  ordered  pairs  of the  form  (b, v)  which  satisfy  the  second  requirement  in  (4)  is  2 × 5 − 1, i.e.  9  while  that  of  ordered  pairs  of  the  type  (c, w)  satisfying  the  third requirement in (4) is 5.  But the ways these three ordered pairs are formed are completely independent of each other.

So the total number of triplets of  ordered pairs  of the  form  {(a, u), (b, v), (c, w)} where  a, b, c, u, v, w  are non-negative integers that satisfy (4) is 5 ×9 ×5 = 225

Thanks and regards,
Kushagra