Age Problem Need Detailed Solutions

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Discussion Overview

The discussion revolves around a mathematical age problem involving the ages of Mark and Amy, specifically focusing on the conditions that their ages multiplied in certain ways yield perfect squares. The problem is framed as a puzzle requiring algebraic manipulation and reasoning.

Discussion Character

  • Mathematical reasoning
  • Homework-related
  • Exploratory

Main Points Raised

  • Some participants define variables for the ages of Mark (M) and Amy (8) and set up equations based on the problem's conditions.
  • One participant expresses confusion about how to proceed after setting up the equations, indicating a perceived dead end.
  • Another participant suggests that the problem may require a trial and error method to find suitable values for Mark's age.
  • One participant proposes that Mark's age must be a perfect square and that another expression involving Mark's age must also be a perfect square.
  • A participant shares their trial and error results, concluding that Mark's age could be 49, providing calculations to support this conclusion.

Areas of Agreement / Disagreement

There is no consensus on a definitive method to solve the problem, with some participants favoring trial and error while others seek a more algebraic approach. Multiple viewpoints on the best method to find Mark's age remain present.

Contextual Notes

Participants express uncertainty about the algebraic steps and the implications of their equations, indicating potential limitations in their approach or understanding of the problem.

Marcelo Arevalo
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The age of Mark now times the age of Amy a year from now is the square of an
integer. The age of Mark a year from now times the age of Amy now is also the
square of an integer. If Amy is 8 years old now, and Mark is now older than 1 but
younger than 100, how old is Mark now?
 
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Marcelo Arevalo said:
The age of Mark now times the age of Amy a year from now is the square of an
integer. The age of Mark a year from now times the age of Amy now is also the
square of an integer. If Amy is 8 years old now, and Mark is now older than 1 but
younger than 100, how old is Mark now?

Write out what you know. Let x be the age of mark and y the age of Amy.

Then x( y + 1) = what? Then what is your other equation?
 
Last edited:
Let : X = Mark age 2 to 99
Y = Amy age = 2

X (Y+1) = \sqrt{A} EQ 1
(X+1) Y = \sqrt{B} EQ 2

Substituting:
X (8+1) = \sqrt{A}
(X+1) 8 = \sqrt{B}

now I am stucked..its a dead end for me.
 
Marcelo Arevalo said:
Let : X = Mark age 2 to 99
Y = Amy age = 2

X (Y+1) = \sqrt{A} EQ 1
(X+1) Y = \sqrt{B} EQ 2

Substituting:
X (8+1) = \sqrt{A}
(X+1) 8 = \sqrt{B}

now I am stucked..its a dead end for me.

Then we have
\begin{align}
x(y+1) &= A\\
x(x + 8 + 1) &= A\\
x^2 + 9x &= A\\
(x +1)y &= B\\
(x + 1)(x + 8) &= B\\
x^2 + 9x + 8 &= B
\end{align}
Is what I get.
 
I would let $M$ be the age of Mark now, and we know Amy is 8. From the problem, we may then state:

$$9M=m^2$$

$$8(M+1)=4(2M+2)=n^2$$

Thus, we know $M$ must be a perfect square and $2M+2$ must also be a perfect square. Can you find such a number $M$?
 
From the statement above. it seems like will be having a Trial & Error method.
Is there another method you can share.

I actually have done it using trial & error which consume most of my time about 2 hours. I arise with the answer of : Mark = 49

what I want to know is real Algebra method, with equations.
hope you could help me. thank you.
 
My trial & Error Method
49 (8+1) = 441 where in \sqrt{441} = 21

(49+1) 8 = 400 where in \sqrt{400} = 20
 

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