Operator Commutation

Computer Science Level 3

$a \times b + c = a + b \times c$

How many ordered triples $(a,b,c)$ of integers with $0 \le a,b,c \le 10$ are there such that the above statement is valid?

Solve this problem with a program.

Inspiration .

The answer is 231.

3 solutions

Arulx Z
Jan 5, 2016

The answer can also be found out manually but here's the code -

1	`sum(a * b + c == a + b * c for a in xrange(11) for b in xrange(11) for c in xrange(11))`

It works because True is also considered as 1 in Python. A small modification can probably greatly speed up the process and allow multiple variables.

Moderator note:

Simple standard approach of running through all possible cases.

There is still no change in efficiency here. As the code runs 1000 times.

Zeeshan Ali - 5 years, 5 months ago

Yes. I have not made any modifications yet. I'll do it later.

Arulx Z - 5 years, 5 months ago

Zeeshan Ali
Jan 5, 2016

Here is an algorithm as one of the approaches:

    count <- 0
    for a <- 0 to 10
        for b <- 0 to 10
            for b <- 0 to 10
                if a*b+c is a+b*c
                    count <- count+1
    return count

I expect even better approach.

Why coding when we can do this with elementary combinatorics?

$a+bc=ab+c\iff (b-1)(a-c)=0$

Now, consider two cases: (i) when $a=c$ and (ii) when $a\neq c$

For case $(i)$ , the LHS is $0$ regardless of what $b$ is, so fixing $a,c$ with the same value $k\textrm{ (say)}$ , we have $11$ choices for $b$ . Since there are also $11$ choices for $k$ , the total number of ordered pairs $(a,b,c)\in\Bbb{Z^3}$ that satisfy the given condition is $11^2=121$ .

For case $(ii)$ , the LHS is $0$ iff $b=1$ . So, fixing $b$ as $1$ , we can take distinct $a,c$ from $11$ choices in $^{11}\mathbf{P}_2=110$ ways.

Obviously, cases $(i)$ and $(ii)$ are mutually exclusive and exhaustive, so the total number of ordered triplets $(a,b,c)\in\Bbb{Z^3}$ that satisfy the problem conditions is $121+110=231$ .

A more general result would be, if $x\leq a,b,c\leq y$ where $x,y\in\Bbb Z$ , then the total number of ordered triples $(a,b,c)\in\Bbb{Z^3}$ that satisfy the given condition is given by $t(2t-1)$ where $t=|y-x+1|$ .

You should easily be able to code this up in Python as follows:

1
2
3

def func(x,y):
    t=abs(y-x+1)
    return t*(2*t-1)

Since $0\leq a,b,c\leq 10$ , calling func(0,10) will give you the required answer.

Prasun Biswas - 5 years, 5 months ago

Nice! Can you add the first part as a solution?

Calvin Lin Staff - 5 years, 5 months ago

Nice! That's exactly what I expected. Thanks for your contribution. Moreover please upload the first part as solution.

Zeeshan Ali - 5 years, 5 months ago

展豪張
Mar 24, 2016

Try to solve it manually:
$ab+c=a+bc$
$(b-1)(a-c)=0$
$b=1$ or $a=c$
By inclusion-exclusion principle, answer = $11^2+11^2-11=231$

Operator Commutation

The answer is 231.

3 solutions

Moderator note:

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