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

#### Abby8100

##### New member

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

##### New member

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

##### Elite Member
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: $$\displaystyle \ \ C + F = \dfrac{3}{2}(B*E)$$

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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.

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#### Dr.Peterson

##### Elite Member
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.
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.

#### Jomo

##### Elite Member
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: $$\displaystyle \ \ C + F = \dfrac{3}{2}(B*E)$$

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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.

#### Jomo

##### Elite Member
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

##### Elite Member
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

##### Elite Member
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

##### New member
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.