What is the fault here?

Find the number of ordered triplets of integers x,y,z  Z+x,y,z\ \in\ Z^+ such that the L.C.M of x,y,z=23×33x,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,zx,y,z are of the form x=2a×3d,y=2b×3e,z=2c×3fx=2^a \times 3^d,y=2^b \times 3^e,z=2^c \times 3^f where a,b,c,d,e,fa,b,c,d,e,f are non negative integers.Now we have that max(a,b,c)=3,max(d,e,f)=3max(a,b,c)=3,max(d,e,f)=3,so at least one of (a,b,c)=3(a,b,c)=3 and same with (d,e,f)(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(0,1,2,3) hence 44 choices,so the answer should be (3×42)2=(48)2=2304(3 \times 4^2)^2=(48)^2=2304,but it isnt,can somebody point out the mistake please.

Note by Adarsh Kumar
3 years, 7 months ago

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Siddhartha is correct; your method involves some double-counting. I looked at this case by case: let jj be the number of elements of (a,b,c)(a,b,c) that equal 33 and kk be the number of elements of (d,e,f)(d,e,f) that equal 33, with both jj and kk ranging from 11 to 33. There are then 33=93*3 = 9 case pairs (j,k)(j,k) to analyze.

For example, for (j,k)=(1,1)(j,k) = (1,1) we can choose one of (a,b,c)(a,b,c) in 33 ways, one of (d,e,f)(d,e,f) in 33 ways, and the remaining four powers can each be any of 0,10,1 or 2.2. This results in a total of 36=7293^{6} = 729 ordered triples (x,y,z)(x,y,z).

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

N(1,1)=36,N(1,2)=N(2,1)=35,N(2,2)=34,N(1,3)=N(3,1)=33,N(2,3)=N(3,2)=32,N(3,3)=1.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 13691369 ordered triples (x,y,z)(x,y,z).

Brian Charlesworth - 3 years, 7 months ago

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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 (4333)2=(37)2=1369 (4^3 - 3^3)^2 = (37)^2 = 1369

Siddhartha Srivastava - 3 years, 7 months ago

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

Somesh Patil - 3 years, 7 months ago

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@Somesh Patil Max(a,b,c) = 3. Therefore each of (a,b,c) (a,b,c) can take values from (0,1,2,3) (0,1,2,3) . This is 43 4^3 .

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

This is equivalent to the cases where values of (a,b,c) (a,b,c) is from 0,1,2 0,1,2 . This is 33 3^3 .

Therefore the required value is 4333 4^3 - 3^3 .We square this since max(d,e,f) =3 also has 4333 4^3 - 3^3 cases

Siddhartha Srivastava - 3 years, 7 months ago

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@Siddhartha Srivastava ohh thanks a lot i get it

Somesh Patil - 3 years, 7 months ago

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Ohk thanx a lot for helping me!

Adarsh Kumar - 3 years, 7 months ago

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

Siddhartha Srivastava - 3 years, 7 months ago

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Thanx for pointing out my mistake.

Adarsh Kumar - 3 years, 7 months ago

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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 - 3 years, 3 months ago

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