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# 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.

1 year, 1 month ago

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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 $$(x,y,z)$$. · 1 year, 1 month 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 $$(4^3 - 3^3)^2 = (37)^2 = 1369$$ · 1 year, 1 month ago

well can u please explain the logic behind your answer? i mean how did you get that expression · 1 year, 1 month ago

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 · 1 year, 1 month ago

ohh thanks a lot i get it · 1 year, 1 month ago

Ohk thanx a lot for helping me! · 1 year, 1 month 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. · 9 months, 3 weeks ago

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