MHB Calculate w/ Defective Calculator: Multiply w/ Add, Subtract, & Reciprocal

AI Thread Summary
The discussion revolves around using a malfunctioning calculator that cannot perform multiplication but can add, subtract, and compute reciprocals. Participants explore methods to effectively multiply two numbers, \(x\) and \(y\), using the available functions. One proposed solution involves calculating \(x+y\) and then manipulating the reciprocals to derive \(xy\). There is some skepticism regarding the necessity of a square root function in one of the solutions, as it may not be available on the defective calculator. Overall, the thread highlights creative mathematical approaches to overcome the limitations of the calculator.
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You have a malfunctioning calculator that cannot perform multiplication. However, it can add, subtract, and compute the reciprocal $\frac{1}{x}$ of any number $x$. Can you nevertheless use this defective calculator to multiply numbers?

Source: 3rd International Young Mathematicians' Convention
 
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Hey Rido! ;)

What? My calculator isn't malfunctioning. It multiplies just fine. (Dull)

Oh well... grumbling... (Sweating)
I'll just do:
$$xy=\frac{x+y}{\frac 1x + \frac 1y}$$ (Whew)
(I recognize the form from parallel resistors. (Smirk))
 
I like Serena said:
Hey Rido! ;)

What? My calculator isn't malfunctioning. It multiplies just fine. (Dull)

Oh well... grumbling... (Sweating)
I'll just do:
$$xy=\frac{x+y}{\frac 1x + \frac 1y}$$ (Whew)
(I recognize the form from parallel resistors. (Smirk))

Very interesting! Thank you for participating. :D However, one point I am not following, and it might be because I don't own a malfunctioning calculator.

You have $x$ and $y$, and so you are able to computed $x+y$ (1) in the numerator. You were also able to take the reciprocal of $\frac{1}{x}$ and $\frac{1}{y}$, and add them to get $\frac{1}{x}+\frac{1}{y}$ (2). But, how were you able to combine step (1) and (2) without being able to multiply? (Wondering) Can you use your defective calculator to confirm?
 
Rido12 said:
Very interesting! Thank you for participating. :D However, one point I am not following, and it might be because I don't own a malfunctioning calculator.

You have $x$ and $y$, and so you are able to computed $x+y$ (1) in the numerator. You were also able to take the reciprocal of $\frac{1}{x}$ and $\frac{1}{y}$, and add them to get $\frac{1}{x}+\frac{1}{y}$ (2). But, how were you able to combine step (1) and (2) without being able to multiply? (Wondering) Can you use your defective calculator to confirm?

Oh darn. I guess I'll have to buy one of those defective calculators just to figure it out. (Crying)
 
Rido12 said:
You have a malfunctioning calculator that cannot perform multiplication. However, it can add, subtract, and compute the reciprocal $\frac{1}{x}$ of any number $x$. Can you nevertheless use this defective calculator to multiply numbers?

[sp]Is...

$\displaystyle x\ y = \frac{x\ y}{x + y}\ (x + y) = \frac{x+y}{\frac{1}{x} + \frac{1}{y}}$

[/sp]

Kind regards

$\chi$ $\sigma$
 
There is a bug (division by zero) in the solutions above if $x = -y$. If this is the case then the alternative version:
$$xy = \frac{xy}{x - y} (x - y) = \frac{x - y}{\frac{1}{y} - \frac{1}{x}}$$
should be used. At least one of the two versions will work except for $x = y = 0$ in which case $xy = 0$ and no calculator is required.
 
Rido12 said:
You have a malfunctioning calculator that cannot perform multiplication. However, it can add, subtract, and compute the reciprocal $\frac{1}{x}$ of any number $x$. Can you nevertheless use this defective calculator to multiply numbers?
[sp]
Step 1.From $x$ you can get $\frac12x$. In fact, $\dfrac1{\frac1x + \frac1x} = \frac12x.$

Step 2. From $x$ you can get $x^2$. In fact, $\dfrac1{x-0.5} - \dfrac1{x+0.5} = \dfrac1{x^2 - 0.25}.$ The reciprocal of that is $x^2 - 0.25$. Add $0.25$ to get $x^2$.

Step 3. From $x$ and $y$ you can get $xy$. In fact, from $x$ and $y$ you can get $x+y$ (by addition), then $(x+y)^2 = x^2 + 2xy + y^2$ (by Step 2). Now subtract $x^2 + y^2$ (which again you can get by Step 2 together with addition) to get $2xy$. Finally, use Step 1 to get $xy.$[/sp]
 
Thanks for participating, Bacterius and $\chi$ $\sigma$, but I am pretty sure you are making an multiplication step in order amalgamate $ (x - y)$ and $\frac{1}{\frac{1}{y} - \frac{1}{x}}$.

@Opalg, excellent solution! (Cool) Thanks for participating.
 
Solution of other:

Solution:
To multiply $x$ by a negative number $-n^2$:
$$-n^2x=n+\frac{1}{-\frac{1}{n}+\frac{1}{n+\frac{1}{x}}}$$

To multiply $x$ by a positive number $n^2$:

$$n^2x=n+\frac{1}{-\frac{n+1}{n}+\frac{1}{1+\frac{1}{-(n+1)+\frac{1}{x}}}}$$
 
  • #10
Rido12 said:
Solution of other:

Solution:
To multiply $x$ by a negative number $-n^2$:
$$-n^2x=n+\frac{1}{-\frac{1}{n}+\frac{1}{n+\frac{1}{x}}}$$

To multiply $x$ by a positive number $n^2$:

$$n^2x=n+\frac{1}{-\frac{n+1}{n}+\frac{1}{1+\frac{1}{-(n+1)+\frac{1}{x}}}}$$
I'm dubious about "other"'s solution. If you want to find $xy$ (with $y$ positive) then this solution says that you need to express $y$ in the form $n^2$. But we are not told whether the calculator has a square root function (or if it has, whether that is also malfunctioning). So how do you find $n$?
 
  • #11
I checked Opalg's solution on my new defective calculator, just to see how it looks: (Thinking)
$$
xy=\frac 1{\frac 1{\frac 1{\frac 1{x+y-\frac 12}-\frac 1{x+y+\frac 12}}
- \frac 1{\frac 1{x-\frac 12}-\frac 1{x+\frac 12}}
- \frac 1{\frac 1{y-\frac 12}-\frac 1{y+\frac 12}}
- \frac 14} + \frac 1{\frac 1{\frac 1{x+y-\frac 12}-\frac 1{x+y+\frac 12}}
- \frac 1{\frac 1{x-\frac 12}-\frac 1{x+\frac 12}}
- \frac 1{\frac 1{y-\frac 12}-\frac 1{y+\frac 12}}
- \frac 14}}
$$
So even on my defective calculator, I can still easily calculate a product. (Happy)
 
  • #12
Hi Opalg, you are absolutely correct about that point. The solution used by the author makes use of the fact that the calculator has a functioning square root button. The author also presented another solution, but that one is identical to the one you posted.

I thought this might be a cool trick to show to friends...or in case of an emergency during a test when the calculator really breaks down...hehe :D

Although the possibility of just the multiplication button not working is just not likely. :p
 

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