Question

Suppose that $a,b,c$ are positive integers such that ${2^a} + {4^b} + {8^c} = 328$ . Then the value of $\dfrac{{a + 2b + 3c}}{{abc}}$ is equal to?
A. $\dfrac{1}{2}$
B. $\dfrac{5}{8}$
C. $\dfrac{{17}}{{24}}$
D. $\dfrac{5}{6}$

Answer 1

Hint: In this question firstly we will split ${4^b},{8^c}$ in such a way that they become powers of 2 i.e. ${2^{2b}},{2^{3c}}$ respectively. Now, our left-hand side of the equation given becomes a series of sum of powers of 2. We know that every number can be written as a sum of some powers of 2. But for this question since our total sum is 328 so we will take powers of 2 up to ${2^8}$ which is 256 because we have to take powers of 2 which are smaller than 328. Now we have to check for the leftover powers for that firstly we will calculate the left-over sum which is $328 - 256 = 72$ . Now we will split the 72 as $8 + 64$ because both can be written as the powers of 2. Now we have ${2^3} + {2^6} + {2^8} = 328$ . Now we have to compare the powers here with a, 2b, 3c and side by side. We have to check which one must be multiple of 3 and which one must be a multiple of 2. After this we will just check for all the values of a, b and c possible to give the answer and then substitute in the expression to have the value and then check with the options which one is correct.

Complete step-by-step answer:

It’s been given that ${2^a} + {4^b} + {8^c} = 328$
$ \Rightarrow {2^a} + {2^{2b}} + {2^{3c}} = 328$

We know that every number can be written as the sum of powers of 2. Just for example, $5 = {2^0} + {2^2}$

But here we will take only that powers of 2 whose value is less than 328 which are \[{2^0},{2^1},{2^2},{2^3},{2^4},{2^5},{2^6},{2^7},{2^8}\]

Let us take ${2^8}$ here because we will need that to make 328.
We know that ${2^8} = 256$
Now we are left with $328 - 256 = 72$

Now to make 72 we will just split 72 into 8 + 64 because 8 can be represented as ${2^3}$ and 64 can be represented as the ${2^6}$
So, we get ${2^3} + {2^6} + {2^8} = 328$

Now on comparing this equation with ${2^a} + {2^{2b}} + {2^{3c}} = 328$ we get to know that one of them must be a multiple of 3 due to 3c. So, 3c can be either 6 or 3. So, c can be either 2 or 1.

Also, one of them must be a multiple of 2 due to 2b. So, 2b can be either 6 or 8. So, b can be either 3 or 4.

The left-over must be our a.

So, the possible values for a, b and c can be
a = 3, b = 4 and c = 2
a = 6, b = 4 and c = 1
a = 8, b = 3 and c = 1

Now, the value of $\dfrac{{a + 2b + 3c}}{{abc}}$ for a = 3, b = 4 and c = 2 is $\dfrac{{3 + 2 \times 4 + 3 \times 2}}{{3 \times 4 \times 2}} = \dfrac{{3 + 8 + 6}}{{24}} = \dfrac{{17}}{{24}}$
value of $\dfrac{{a + 2b + 3c}}{{abc}}$ for a =6, b = 4 and c = 1 is $\dfrac{{6 + 2 \times 4 + 3 \times 1}}{{6 \times 4 \times 1}} = \dfrac{{6 + 8 + 3}}{{24}} = \dfrac{{17}}{{24}}$
and the value of $\dfrac{{a + 2b + 3c}}{{abc}}$ for a = 8, b = 3 and c = 1 is $\dfrac{{8 + 2 \times 3 + 3 \times 1}}{{8 \times 3 \times 1}} = \dfrac{{8 + 6 + 3}}{{24}} = \dfrac{{17}}{{24}}$

We get to know that for all the cases, the value of $\dfrac{{a + 2b + 3c}}{{abc}}$ comes out to be $\dfrac{{17}}{{24}}$

Hence, the value of $\dfrac{{a + 2b + 3c}}{{abc}}$ is $\dfrac{{17}}{{24}}$

$\therefore $ Option C. $\dfrac{{17}}{{24}}$ is our correct answer.

Note: We can also solve this question rewriting the equation ${2^a} + {4^b} + {8^c} = 328$ as ${2^a}\left( {1 + {2^{2b - a}} + {2^{3c - a}}} \right) = {2^3} \times 41$ . On comparing we get to know a = 3. Since 41 is a prime number we can’t split it further. We will subtract 1 from 41 and \[\left( {1 + {2^{2b - a}} + {2^{3c - a}}} \right)\] so that we get $\left( {{2^{2b - a}} + {2^{3c - a}}} \right) = 40$ , which can also be written as $\left( {{2^{2b - a}} + {2^{3c - a}}} \right) = 5 \times {2^3}$ .On comparing we get, 3c – a = 3 and 2b – 3c = 2 due to which we get a = 3, b = 4, c = 2 on simplifying the equations. On substituting the values of a = 3, b = 4 and c = 2 in the $\dfrac{{a + 2b + 3c}}{{abc}}$ we get $\dfrac{{3 + 2 \times 4 + 3 \times 2}}{{3 \times 4 \times 2}} = \dfrac{{17}}{{24}}$ . This is another approach to this problem.