Waste less time on Facebook — follow Brilliant.
×

What is the fault here?

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.

Note by Adarsh Kumar
11 months ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

Siddhartha 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 · 11 months ago

Log in to reply

@Brian Charlesworth 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 \( (4^3 - 3^3)^2 = (37)^2 = 1369 \) Siddhartha Srivastava · 11 months ago

Log in to reply

@Siddhartha Srivastava well can u please explain the logic behind your answer? i mean how did you get that expression Somesh Patil · 11 months ago

Log in to reply

@Somesh Patil Max(a,b,c) = 3. Therefore each of \( (a,b,c) \) can take values from \( (0,1,2,3) \). This is \( 4^3 \).

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 · 11 months ago

Log in to reply

@Siddhartha Srivastava ohh thanks a lot i get it Somesh Patil · 11 months ago

Log in to reply

@Brian Charlesworth Ohk thanx a lot for helping me! Adarsh Kumar · 11 months ago

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 · 7 months ago

Log in to reply

You're double counting many cases. For example, you count the case (3,3,0) two times. Siddhartha Srivastava · 11 months ago

Log in to reply

@Siddhartha Srivastava Thanx for pointing out my mistake. Adarsh Kumar · 11 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...