Find the number of ordered triplets of integers \(x,y,z\ \in\ Z^+\) such that the L.C.M of \(x,y,z=2^3\times 3^3\)?\[\] My approach:

Since we are talking about the lease common multiple we can safely assume that all \(x,y,z\) are of the form \(x=2^a \times 3^d,y=2^b \times 3^e,z=2^c \times 3^f\) where \(a,b,c,d,e,f\) are non negative integers.Now we have that \(max(a,b,c)=3,max(d,e,f)=3\),so at least one of \((a,b,c)=3\) and same with \((d,e,f)\),now there are three ways of choosing that one,and after that is done the two left can be \((0,1,2,3\)) hence \(4\) choices,so the answer should be \((3 \times 4^2)^2=(48)^2=2304\),but it isnt,can somebody point out the mistake please.

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TopNewestSiddhartha is correct; your method involves some double-counting. I looked at this case by case: let \(j\) be the number of elements of \((a,b,c)\) that equal \(3\) and \(k\) be the number of elements of \((d,e,f)\) that equal \(3\), with both \(j\) and \(k\) ranging from \(1\) to \(3\). There are then \(3*3 = 9\) case pairs \((j,k)\) to analyze.

For example, for \((j,k) = (1,1)\) we can choose one of \((a,b,c)\) in \(3\) ways, one of \((d,e,f)\) in \(3\) ways, and the remaining four powers can each be any of \(0,1\) or \(2.\) This results in a total of \(3^{6} = 729\) ordered triples \((x,y,z)\).

With \(N(j,k)\) being the number of ordered triples \((x,y,z)\) resulting from particular values of \((j,k)\), I find that

\(N(1,1) = 3^{6}, N(1,2) = N(2,1) = 3^{5}, N(2,2) = 3^{4}, N(1,3) = N(3,1) = 3^{3}, N(2,3) = N(3,2) = 3^{2}, N(3,3) = 1.\)

These values sum to a total of \(1369\) ordered triples \((x,y,z)\). – Brian Charlesworth · 1 year ago

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– Siddhartha Srivastava · 1 year ago

A simpler solution is the square of the number of ways to choose (a,b,c) from (0,1,2,3) minus the number of ways to choose (a,b,c) from (0,1,2). This is \( (4^3 - 3^3)^2 = (37)^2 = 1369 \)Log in to reply

– Somesh Patil · 1 year ago

well can u please explain the logic behind your answer? i mean how did you get that expressionLog in to reply

Now, we subtract cases where none of \( (a,b,c) \) are \( 3 \).

This is equivalent to the cases where values of \( (a,b,c) \) is from \( 0,1,2 \). This is \( 3^3 \).

Therefore the required value is \( 4^3 - 3^3 \).We square this since max(d,e,f) =3 also has \( 4^3 - 3^3 \) cases – Siddhartha Srivastava · 1 year ago

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– Somesh Patil · 1 year ago

ohh thanks a lot i get itLog in to reply

– Adarsh Kumar · 1 year ago

Ohk thanx a lot for helping me!Log in to reply

Since we are taking LCM so we can assume x = 2^a 3^d, y = 2^b 3^e, z = 2^c 3^f, where a, b, c, d, e, f, g are non negative integers. As per question max {a, b, c} = 3 and max {d, e, f} = 3. Number of ways in which max {a, b, c} = 3 is 4^3 – 3^3 = 37. Similarly, number of cases for max {d, e, f} = 3 is 37. Hence numbers of required ordered triplets = 37 × 37 = 1369. – Ramesh Chandra · 8 months, 1 week ago

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You're double counting many cases. For example, you count the case (3,3,0) two times. – Siddhartha Srivastava · 1 year ago

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– Adarsh Kumar · 1 year ago

Thanx for pointing out my mistake.Log in to reply