Let $T$ be the following set of ordered triplets,$T=\{(a,b,c):a,b,c\in N\}$. Find the number of elements in $T$ such that $L.C.M(a,b,c)=72$.

algebra-precalculuscombinationscombinatoricselementary-number-theorypermutations

My Attempt
Let $a=2^{x_1}3^{y_1}$,$b=2^{x_2}3^{y_2}$ and $c=2^{x_3}3^{y_3}$

The $x_i's$ have 4 choices each where at least one of them should have the value $3$
and the rest can take any value from $0$ to $4$. This can be done in $3\times 4\times 4=48$ ways

Similarly for $y_i's$ we have $3$ choices for each. This can be done in $27$ ways .

So required number of ways =$48\times 27=1245$ ways.

Is this correct

Best Answer

There are $3$ possible values for the exponent of $3$ in $A,B,C$ but at least one must take the maximal value of $2$. So there are $3^3 - 2^3 = 27-8 = 19$ possibilities.

Similarly for the exponents of $2$ there are $4^3 - 3^3 = 37$ possibilities.

$37*19 = 703$

Related Solutions

[Math] Find the number of integer solutions for $x_1+x_2+x_3 = 15$ under some constraints by IEP.

Here is a different way to break it down $$ x_1\in\{1,2,3,4,5\} $$ and given $x_1$ we then have $x_1+x_2<15$ and $x_2>6$ combined as $$ 6<x_2<15-x_1 $$ And whenever $x_1$ and $x_2$ are given, the value of $x_3$ follows from them.

For $x_1=5$ we then have $x_2\in\{7,8,9\}$ so three choices for $x_2$. Each time $x_1$ is decreased by $1$ we gain one option for $x_2$. Thus we have a total of $$ 3+4+5+6+7 = 25 $$ sets of integer solutions under the given constraints.

I ran the following code snippet in Python which confirmed the figure of 25:

n = 0
for x1 in range(1,16):
    for x2 in range(1,16):
        for x3 in range(1,16):
            if x1 < 6 and x2 > 6 and x1+x2+x3 == 15:
                n += 1
                print n, ":", x1, x2, x3

I understand that I did not answer the question using the method required, but I wonder why I find the number of solutions to be $25$ whereas the OP and the other answer find it to be $49$. Did I misunderstand the question in the first place?

Find the number of ordered triplets $(a, b, c)$ of positive integers such that $30a + 50b + 70c < 343$

The problem is that if $c=3$ you need $30a+50b \lt 133$. In that case $b$ can only be $1$ or $2$. If $b=2$ you must have $a=1$, while if $b=1$ you can have $a=1$ or $2$. The choices are not independent, so you cannot multiply.

You can just continue the casework. There are not too many cases for $b,c$.

Best Answer

Related Solutions

[Math] Find the number of integer solutions for $x_1+x_2+x_3 = 15$ under some constraints by IEP.

Find the number of ordered triplets $(a, b, c)$ of positive integers such that $30a + 50b + 70c < 343$

Related Question