Multiply a number by x using only addition. Solved the question, but want to understand how it works?

[LukeSwift]

The first problem was 'multiply a number by 8 using only addition'.

Number 2

2 x 8 = 16

Solution:
2 + 2 = 4
4 + 4 = 8
8 + 8 = 16

The reason I really knew how to do it was the game (Human Resources Machine), asked 'can you solve this question using only three addition steps'? So that gave me a clue to simply add the results of each addition step together, which ended up being the correct answer.

My question is why/how does this work?

The next question was to multiply a number by 40.

Number 2

2+2 = 4
4+4 = 8
8+2 = 10
10 + 10 = 20
20 + 20 = 40

Again here I simply experiment around and got the answer - really from the constraints of only having an addition operator, and just trial and error.

My question is ultimately, to understand why this works (in both cases) so that I can apply the rule in the future to other problems.

Many thanks,
Lukess

 

[LukeSwift]

**Sorry I thought my other post didn't post correctly**


Number 2
x = 8

2 x 8 = 16

2+2 = 4
4+4 = 8
8+8 = 16

The game I am playing (Human Resource Machine) give me a hint, 'use only three addition operations', so I teased the above out. However, why does adding like this give the answer? I know it does hold through for all numbers, for example, the next question was to multiply the number by 40.

Number 2
x = 40

2+2 = 4
4+4 = 8
2+8 = 10
10+10 = 20
20+20 = 40
40+40 = 80

So again, I found this by doing a bit of trial and error, however, I would like to know what it is I'm doing that makes this works, what is the principle underneath I am using to get the correct answer?

Is it something to do with 'the power of', perhaps I just don't understand this mathematically?

Many thanks,
Mark

 

[HallsofIvy]

As long as x is a positive integer, xa is equal to "a added to itself x times". Is that what you are talking about?

 

[Dr.Peterson]

The first idea here is that adding a number to itself is the same as multiplying it by 2. As a result, we can multiply by any power of two by repeated addition to self (doubling).

Multiplying by 8, which is 2*2*2, can be done by repeating that process three times, i.e. adding a number to itself three times:

y = x * 2 = x + x​

z = x * 4 = x * 2 * 2 = y * 2 = y + y​

x * 8 = x * 4 * 2 = z*2 = z + z​

The second idea is that any number can be written as a sum of powers of 2; this is related to the binary number system.

Multiplying by 5, which is 4 + 1, can be done by multiplying by 4 and by 1 and adding:

x * 40 = x * 5 * 8 = x * (4 + 1) * 8 = (x * 4 + x * 1) * 8 = ...​

 

[LukeSwift]

As long as x is a positive integer, xa is equal to "a added to itself x times". Is that what you are talking about?

The first idea here is that adding a number to itself is the same as multiplying it by 2. As a result, we can multiply by any power of two by repeated addition to self (doubling).

Ok, I see, it's very basic when you spell it out. I think this rule 'we can multiply by any power of two by repeated addition to self' is what I'm latching onto, and what I used in both solutions.

The programming challenge said I could only use addition operators, even though it asked me to multiply a number by, 51 say. I can use my initial number at any time to add with, and I could use two other results from my addition results, to add with also (only three numbers can be retained in computer memory.)

Number = 2
51 x 2 = 102

So how do I get 102 from only addition operations?

From what I've learned, I can go 'laterally' (2 + 2 = 4) using my power of two rule, and then I can go 'horizontally', by simply adding my initial number with the result of my first calculation, 4, so 2 + 4 = 6. I can then either go horizontally again, by adding either 2, 4, or 6 to each other, or I could go laterally again by 6+6 = 12 (6^2).

I would keep performing some combination of the above two operations until I get to 102.

Does this make sense?

Is there a better/quicker way to multiply a number by x, without actually using multiplication?

Are the above 'lateral', and 'horizontal' operations, the only moves I have available with all I have an addition operator?

Many thanks for your responses.

Lukess

 

[Dr.Peterson]

I'm not sure you have yet stated the entire problem exactly as given to you, as we ask in our submission guidelines. It's not clear whether this is a "programming challenge" for which you have to write a general program in some language, or just a game asking you to find an expression for a particular computation, or something else. Please tell us all about it.

If you are saying here that you have to write a single expression the yields 102, with only the number 2 and three addition operators, I believe that's impossible.

 

[lev888]

If you have a programming challenge I'm assuming you need to write a program that multiplies input numbers using only addition. Looks like a good use case for the loop control statement.
Or is the problem to come up with fewest additions to get the given result?
Please clarify.