Product is Thrice Sum

Number Theory Level pending

How many ordered sets of positive integers $(x, y, z)$ are there such that $x \times y \times z= 3(x+y+z)?$

The answer is 28.

2 solutions

Calvin Lin Staff
May 13, 2014

Since 3 divides the RHS, it has to divide the LHS. Without loss of generality, we may assume that $x$ is a multiple of 3. Let $x=3w$ , where $w$ is a positive integer. Making $w$ the subject of the equation gives $w = \frac {y+z}{yz-3}$ . Without loss of generality, we let $z \leq y$ . We consider the following cases:

Case 1: $z=1$ . Then $w = \frac {y+1}{y-3} = 1 + \frac {4}{y-3}$ , so $y-3$ must be a divisor of 4, so $y-3 = 1, 2, 4$ which gives $y = 4,5, 7$ . This yields solutions $(x, y, z) = (15, 4, 1), (9, 5, 1), (6, 7, 1)$ . Without restriction on $x, y$ and $z$ , there will be $3 \cdot 3! = 18$ distinct ordered sets of solution.

Case 2: $z > 1$ . Since $w \geq 1$ , we have $y+z \geq yz -3 \Rightarrow (y-1)(z-1) \leq 4$ . Since $y-1$ and $z-1$ are positive integers, this greatly restricts the possibilities of $(y, z)$ to $(2, 2), (3, 2), (4, 2), (5, 2)$ and $(3, 3)$ . We then test each case to get an integer solution to $w$ , and find that the only solutions are $(x, y, z) = (12, 2, 2), (3, 5, 2), (3, 3, 3)$ . Without restriction on $x, y$ and $z$ , the first set yields 3 distinct ordered solutions, the second set yields $3! = 6$ distinct ordered solutions, and the last set yields 1 distinct ordered solution.

Thus, there are $18+3+6+1= 28$ distinct ordered solutions in all.

Sreejato Bhattacharya
May 20, 2014

By symmetry assume that x>=y>=z.

So, 3(x+y+z)<=3(x+x+x)= 9x

Or, 3xyz<=9x Or, yz<=9 [since by condition a is strictly positive]

So the possible values of bc are 1, 2, 3, 4, 5, 6, 7, 8, & 9.

The following chart is made.

Value of yz Possible values of y and z Corresponding value of x No. of permutations of (x, y, z)
1 b=1 c=1 a= -3 [Not possible] Out of context 2 b=2 c=1 a= -9 [Not possible] Out of context 3 b=3 c=1 No solution Out of context 4 b=4 c=1 a= 15 3!= 6 4 b=2 c=2 a= 12 3!= 6 5 b=5 c=1 a= 9 3!= 6 6 b=6 c=1 a= 7 3!= 6 6 b=3 c=2 a= 5 3!= 6 7 b=7 c=1 a= 6 3!= 6 8 b=8 c=1 a= 5.4 [Not possible] Out of context
8 b=4 c=2 a= 6 3!= 6 9 b=9 c=1 a= 5 3!= 6 9 b=3 c=3 a= 3 3!= 6

The number of permutations is considered because if (x, y, z) is a solution, so is every permutation of (x, y, z).

Adding the numbers, we get it to be 6+6+6+6+6+6+6+6+6= 9*6= 45

So, the answer is 45.

Comments and replies:

Sreejato Bhattacharya:

In the table I wrote a,b, c in place of x, y, z by mistake.

The table has been corrected and arranged in the following link.

http://postimage.org/image/ch9f1jo45/

[Table corrected and link updated]

Calvin:

I like the initial step of bounding $yz$ so that we only need to consider a few cases.

I'm slightly confused. It appears that you switched notation from $x, y, z$ to $a, b, c$ . I would like to do a direct substitution that $x=a, y=b, z=c$ . However, you also had an initial condition that $x \geq y \geq z$ , and I do not see a corresponding requirement that $a \geq b \geq c$ . (for example, you have $(a, b, c) = (6, 7, 1)$ , when value of $yz=7$ . Can you explain what is going on here?

Also, Can you explain how, given a solution $(x, y, z)$ , to calculate the total number of permutations of $(x, y, z)$ ? Pay particular attention to your sets $(a, b, c) = (15, 4, 1), (12, 2, 2)$ and $(3, 3, 3)$

Calvin:

The solution is short enough, and your cases are simply stated.

You have not addressed my concern that you initially assumed that \( x \geq y \geq z$ and didn't verify that this condition still held in your solutions.

Also, you have double counted several solutions. After listing out all the solutions, you still have to compare them to see that you obtained distinct sets of solutions. Otherwise, I can list \( (3, 3, 3)$ as an answer 200 times to get "200 solutions".

[Note: This comment refers to the case (1,6,7), which was later identified as a repeat solution and the table was updated.]

-----------------------------

Sreejato Bhattacharya:

Very sorry for switching notations from x, y, z to a,b, c. The table has been corrected and uploaded in http://postimage.org/image/ch9f1jo45/.

If x, y, z are distinct, they can be permuted in 3!= 6 ways. If one of x, y, z are equal, they can be permuted in 3!/2!1!= 3 ways. If x=y=z, they can permuted in 1 way.

Adding the numbers, we get it to be 6+6+6+6+6+3+1= 34.

I think that the problem could be made shorter by not considering the cases when yz= 1, yz= 2, and yz= 3.

This is because $ abc= 3(a+b+c)$ implies that $a(bc-3)= 3(b+c)$ .

Since both $a$ and $3(b+c)$ are positive, so must be $bc- 3$ .

So...

$bc- 3> 0$ Or, $bc>3$

Therefore...

$9\geq bc>3$

The cases for yz= 1, yz= 2, and yz= 3 were not necessary.

[Latex typo corrected.]

In the table [http://postimage.org/image/ch9f1jo45/.] I have verified which sets of solutions are actually permutations of other sets of solutions.

The [distinct] solution sets are listed below.

a) (x, y, z)= (4, 1, 15) b) (x, y, z)= (2, 2, 12) c) (x, y, z)= (5, 1, 9) d) (x, y, z)= (6, 1, 7) e) (x, y, z)= (3, 2, 5) f) (x, y, z)= (4, 2, 6) g) (x, y, z)= (3, 3, 3)

Note:- If $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are two solution sets where $(x_2, y_2, z_2)$ is a permutation of $(x_1, y_1, z_1)$ , only one of them is listed.

No. of permutations of set(a)= 3!= 6 No. of permutations of set(b)= 3!/(2!1!)= 3 No. of permutations of set(c)= 3!= 6 No. of permutations of set(d)= 3!= 6 No. of permutations of set(e)= 3!= 6 No. of permutations of set(f)= 3!= 6 No. of permutations of set(g)= 1

The number of permutations is considered because if $(x_1, y_1, z_1)$ is a solution set, so must be every permutation of $(x_1, y_1, z_1)$ .

Adding these, the total number is 6+6+6+6+6+3+1= 34

Calvin:

Please check your arithmetic of all your cases.

Sreejato Bhattacharya:

set(a)...

4+1+15= 20 4 1 15= 60

This is true.

set(b)...

2+2+12= 16 2 2 12= 48

This is true.

set (c)...

5+1+9= 15 5 9 1= 45

This is true.

set(d)...

6+1+7= 14 6 7 1= 42

This is true.

set(e)...

3+2+5= 10 3 2 5= 30

This is true.

set(f)

4+2+6= 12 4 2 6= 48

This is false.

It should have been... (x, y, z)= (4, 2, 3.6) [impossible]

So, this is out of context.

set(g)

3+3+3= 9 3 3 3= 27

This is true.

Thus, set(f) is cancelled.

now adding the total number of permutations...

6+6+6+6+1+3= 28

Sorry for the mistake.

[Edits for set (b)]

Calvin:

This now looks good. Great job.

Sreejato Bhattachaya:

set(b) is right, but its calculations ae wrong.

2+2+12= 16 2 2 16= 48

Calvin Lin Staff - 7 years ago

Product is Thrice Sum

The answer is 28.

2 solutions

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