A year hidden between the powers of 2!

$\Large 2^a-2^b=2016$

Since $a>b$ , then , $\large 2^b(2^{a-b}-1)=2016=2^5\times 3^2 \times 7$ .

But since $(2^{a-b}-1)$ is always odd , no power of $\large 2$ must be provided to it.Thus by comparing $L.H.S$ and $R.H.S$ we get :

$\large \rightarrow 2^b=2^5 \Rightarrow \boxed{b=5} \\ \large\rightarrow 2^{a-b}-1 = 63 \Rightarrow 2^{a-5}=64=2^6 \Rightarrow a-5=6 \Rightarrow \boxed{a=11}$

Thus , $\large a+b=11+5=\boxed{16}$

Question:

$2^{a}-2^{b}=2016 \; (1)$

I am an Engineer, so unlike mathematicians, we start every solution by making some assumptions based on the evidence before us. First I assume that if someone posted this question to Brilliant, then there is one unique solution that will result in the screen turning green when I hit submit. After that, things get quite simple.

If 2016 is converted to binary, the solution is clear by inspection.

$2016 = 1111100000_{2}$

In binary, the solutions for $2^{a}$ and $2^{b}$ take the form of a $1$ followed by $a$ and $b$ zeros respectively. Finding two such numbers to create a difference resulting in $1111100000_{2}$ gives the solution to eqn. (1) expressed in binary.

$10000000000_{2}-10000 = 1111100000_{2}$

Converting back to base 10 we see eqn. (1) in a more familiar form.

$2^{11}-2^{5} = 2016\\ a = 11\\ b = 5$

The solution is the sum of a and b.

$a+b = 16\\$

2016 = 2 ^5 * 63 ....... 64 - 1 = 63 .......... 64 = 2^6 .........
2^5( 2^6 - 1 ) = 2 ^ 11 - 2 ^ 5 for this reason : a + b = 16

Since 2^11= 2048 and 2048-2016=32 and 2^5 is 32 and 11+5=16 We use 2048 because 2048 is the closest to 2016

Nearest integer greater than 2016 which is power of 12 is - 2048. Now, 2048-2016 = 32. 32 = 2^5. 2048 = 2^11. 2016 = 2048 - 32. 2016 = 2^11 - 2^5.

11+5=16. Answer is 16.

By trials,we put a=11 then 2^11=2048,b=5 so 2^5=32,expression=2048-32=2016,,a+b=16

2048 - 32 2∧11_2∧5

$2^a - 2^b = 2016$

Noticing that there are powers of 2 on the L.H.S. and 2016 is an even number, I wanted to see if I could manipulate the R.H.S. to become a power of 2. Then by inspection, we would be able to see what the answer was clearly. Here is the method I used:

Prime Factorize 2016:

$2016 = 2^5\times3^2\times7$

Divide both sides by $2^5$ and multiply $3^2$ and 7 on the R.H.S. to get 63. The equation will look like this:

$\dfrac {2^a - 2^b}{2^5} = 63$

Notice that the 63 is only 1 away from a power of 2.

64 is $2^6$ . If I add $1$ to both sides, I will get:

$\dfrac {2^a - 2^b}{2^5}+1 = 63+1$

$\dfrac {2^a - 2^b}{2^5}+1 = 64$

Rewriting 1 and 64 as powers of 2:

$\dfrac {2^a - 2^b}{2^5}+2^0 = 2^6$

Subtracting $2^0$ from both sides:

$\dfrac {2^a - 2^b}{2^5} = 2^6 - 2^0$

Multiplying both sides by $2^5$ :

$2^a - 2^b = 2^5(2^6 - 2^0)$

Distributing:

$2^a - 2^b = 2^{11} - 2^5$

Here we can see what $a$ and $b$ are:

$a = 11$ and $b = 5$

Therefore $a+b = 16$

I I'm pretty adept in base 2 since I'm a computer science enthusiast so I knew 2^11 = 2048 and I knew 2^5 = 32 so I just added the exponent and I got 16

(2^a - 2^b) = (2^11 - 2^5) So a=11 b=5

This is equal to 2048-32 which gives us the answer