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.

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Comments

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 - 5 years, 5 months 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 $(4^3 - 3^3)^2 = (37)^2 = 1369$

Siddhartha Srivastava - 5 years, 5 months ago

well can u please explain the logic behind your answer? i mean how did you get that expression

Somesh Patil - 5 years, 5 months ago

@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 - 5 years, 5 months ago

@Siddhartha Srivastava – ohh thanks a lot i get it

Somesh Patil - 5 years, 5 months ago

Ohk thanx a lot for helping me!

Adarsh Kumar - 5 years, 5 months ago

You're double counting many cases. For example, you count the case (3,3,0) two times.

Siddhartha Srivastava - 5 years, 5 months ago

Thanx for pointing out my mistake.

Adarsh Kumar - 5 years, 5 months ago

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 - 5 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$