Is there an equation that can be written to solve this type of problem or is it guess and check?

[Abby8100]

I am having trouble solving this- would you please show me how you go about thinking about this. Where do you start? Thank you

Question: Each student has two numbers from 1,2,3,4,5 6. Which two numbers does each student have?

Allison: We each have two numbers. The sum of my numbers is one fourth the product of Charma's tiles.

Charma: The sum of my numbers is one and a half times the product of Bashar's numbers.

Bashar: the product of my numbers is equal to the product of Allison's numbers.

 

[Abby8100]

I am having trouble solving this- would you please show me how you go about thinking about this. Where do you start? Thank you

Question: Each student has two numbers from 1,2,3,4,5 6. Which two numbers does each student have?

Allison: We each have two numbers. The sum of my numbers is one fourth the product of Charma's tiles.

Charma: The sum of my numbers is one and a half times the product of Bashar's numbers.

Bashar: the product of my numbers is equal to the product of Allison's numbers.

 

[lookagain]

Question: Each student has two numbers from 1,2,3,4,5 6. Which two numbers does each student have?

Allison: We each have two numbers. The sum of my numbers is one fourth the product of Charma's tiles.

Charma: The sum of my numbers is one and a half times the product of Bashar's numbers.

Bashar: the product of my numbers is equal to the product of Allison's numbers.

I put the information in a partially different order:

Let Allison's numbers = A , D
Let Bashar's numbers = B , E
Let Charma's number's = C , F


One and a half is the same as \(\displaystyle \ \tfrac{3}{2}.\)


Allison: \(\displaystyle \ \ A + D = \ \dfrac{1}{4}(C*F)\)

Bashar: \(\displaystyle \ \ B*E = A*D \)

Charma: \(\displaystyle \ \ C + F = \dfrac{3}{2}(B*E)\)


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


I would look at the conditions of Bashar's line first. Look at the list of six numbers to choose from to see
what B*E = A*D must be.

 

[Dr.Peterson]

Is there an equation that can be written to solve this type of problem or is it guess and check?

You've seen the equations; but I wouldn't put much effort into solving them. They mostly help focus your attention.

I am having trouble solving this- would you please show me how you go about thinking about this. Where do you start? Thank you

Since there are three equations, six unknowns, and an extra condition (that the numbers are 1, 2, 3, 4, 5, 6 in some order), this will require guessing and checking, but not many errors! I would start with Bashar's fact, which tells you a lot. Then the other facts will help you arrange the pairs.

 

[Steven G]

I put the information in a partially different order:

Let Allison's numbers = A + D
Let Bashar's numbers = B + E
Let Charma's number's = C + F


One and a half is the same as \(\displaystyle \ \tfrac{3}{2}.\)


Allison: \(\displaystyle \ \ A + D = \ \dfrac{1}{4}(C*F)\)

Bashar: \(\displaystyle \ \ B*E = A*D \)

Charma: \(\displaystyle \ \ C + F = \dfrac{3}{2}(B*E)\)


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


I would look at the conditions of Bashar's line first. Look at the list of six numbers to choose from to see
what B*E = A*D must be.

You wrote Let Allison'e numbers = A+D while A+D is just one number. You meant to write Let Allison's numbers be A and D. Same problem for the other two statements.

 

[Steven G]

C+F = 3BE/2.

Note that C+F < 11

So 3BE/2 < 11 and BE must be even. That is, 3BE < 22 which actually implies that 3BE < 21 or BE < 7

If B = 1, then E = 2, 4 or 6
If B = 2, then E = 1, 2 or 3
If B = 3, then E = 2
If B = 4, then E = ?
If B = 5, then E = ?
If B = 6, then E = ?

What can C and F be given the values of B and E?

 

[lookagain]

You wrote Let Allison'e numbers = A+D while A+D is just one number. You meant to write . . .

My content is wrong. Thank you for the correction.

By the way, I spelled "Allison's numbers" correctly in post #2.

 

[Otis]

A+D is just one number

Yes, but I find it interesting (enough) to note that A+D may be read as "A and D". We can't always fault our unconscious mind!

 

[StarFriedTree]

I don't know about forming an equation. just thinking about it a little is sufficient.
for convenience, Allison is A, Charma is C, and Bashar is B
A's numbers are A₁ and A₂, B's are B₁ and B₂, and C's are C₁ and C₂

A₁ = 2 , A₂ = 3
as 2+3=5 which is one fourth of 20
20 can be gotten by 5x4 so
C₁ = 4 and C₂ = 5
remaining is 1 and 6.
so, B₁ = 1 and B₂ = 6

you can check these answers match all other conditions of the question too.

 

[Dr.Peterson]

I am having trouble solving this- would you please show me how you go about thinking about this. Where do you start? Thank you

Question: Each student has two numbers from 1,2,3,4,5 6. Which two numbers does each student have?

Allison: We each have two numbers. The sum of my numbers is one fourth the product of Charma's tiles.

Charma: The sum of my numbers is one and a half times the product of Bashar's numbers.

Bashar: the product of my numbers is equal to the product of Allison's numbers.

You could write three equations, but that wouldn't be enough to solve algebraically. The key idea is that the six numbers are distinct numbers from the given set, which an equation can't deal with.

I would start with Bashar's statement, as it involves two products. The only distinct pairs that give the same product are 1*6 and 2*3, or 2*6 and 3*4, so then you just have to try each of those pairs for Allison and for Charma.