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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. 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.
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.
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 AskiitiansExpert-IIT Delhi 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. Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar..
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
AskiitiansExpert-IIT Delhi
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.
Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar..
Dear student,Please find the answer to your problem below. 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
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