Wonderful Algebra Question | Equation Solving | You should learn this method

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How to solve this equation? Deal it quickly by using this trick!

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As a programmer used to work with powers of two, I instantly recognised this as 2048 - 32, or 2¹¹ - 2⁵. I know powers of two by memory up to 2²⁰, plus a few other important ones.

nineko

Easier:
2ᵃ must be the smallest power of 2 greater than 2016. That would be 2¹¹ (2048). Picking any larger power of 2 for 2ᵃ would result in 2ᵃ - 2ᵇ being too high.
For example, if we pick a=12, and then b=11, we get 2¹² - 2¹¹ = 2048 (too big) and any other choice for b results in an even bigger difference; no way to get down to 2016.
So a is 11. Now we simply subtract 2¹¹ - 2016 = 32, which is 2⁵. So b is 5.

kenhaley

How do you determine what to turn 2016 into. Why not 7 x 288 or 3 x 672

goodshowolboy

I solved this before watching the video. I took the hints from powers of 2 and was thinking about converting 2016 to binary, when I noticed that 2016 is the difference between 2048 and 32. OK, so I stumbled upon a and b for this case. But how to solve this in general?

First off, the difference must be a number of the form (2^x)(2^y - 1). When you convert such a number to binary, you get a representation with y 1s, followed by x 0s. (For 2016, that's six 1s, five 0s.)

The value of b is the number of 0s in the binary form of the difference. In this case, it's 5, since 2^5 is what you'd need to add to to get a binary value having a 1 followed by all 0s.

ricktoews

Quick method. 2016=2^5 x 63. So divide by 2^5 to get: 2^(a-5)-2^(b-5)=63 an odd number. This implies 2^(b-5)=1 and it follows 2^(a-5)=64. No substitutions, no thinking of powers of 2.

PeterMcDaid

Given you're looking for a, b ∈ ℕ, we know 2^a and 2^b are positive integers, so a > b. Taking the highest powers of 2 out of each side: (2^b) * (2^(a-b) - 1) = 32 * 63.
So the two positive integers on the LHS are (a power of 2) and (1 less than a power of 2). The only way the RHS can match that pattern is 2^b = 32 and (2^(a-b) - 1) = 63. So b=5 and a = 6+5 = 11.

RexxSchneider

2^a - 2016 = 2^b;
2^5 * (2^(a - 5) - 63) = 2^b;
2^(a - 5) - 63 can only be odd, because it is (even - odd = odd)
and an odd number can not be a power of two except one special value 1 = 2^0;
2^5 = 2^b; b = 5;
2^(a - 5) - 63 = 1; 2^(a - 5) = 64; a - 5 = 6; a = 11.

nikolaymatveychuk

Thank you for the fascinating and educational videos!

gennoveus

How do I make my handwriting this good? :(

ayan

easy instant visual left decomposition in power of 2 and a, b matches: 2^a-2^b=2016=2048-32=2^11-2^5 >> a=11 b=5 😆

EmmanuelBrandt

Thought the symbol for integers was Z not N?

luygvkz

My solution was easier, you just calculate 2^X for x = 0 to 12, anything over that is too large. And just find 2 numbers that subtract to give 2016. One has to be 2048 (or 2^11) as other numbers are too large or too small by visual inspection. And then you see 32 is 2^5 and 2048 - 32 = 2016. It's easy to calculate powers of 2 up to 12. So no need to do any algebra.

YTSparty

Are there any other solutions to this equation?

varshinid

2048=2^11
32=2^5
a=11, b=32
Any positive integer solution to reach a solution in any of these cases never needs to consider beyond the power that is just above the number reached.

RoderickEtheria

Here it can be seen that 2016 is a difference of two different powers of 2. So let's try to find a way to represent 2016 as a difference of two different powers of two. For simplifying that further, we can divide 2016 by 2 untill an odd number comes.

2016 = (2)(1008)
=(2)(2)(504)
=(2)(2)(2)(252)
=(2)(2)(2)(2)(126)
=(2)(2)(2)(2)(2)(63)
=(32)(63)
=(2^5)(64-1)
=(2^5)(2^6 - 1)
=2^11 - 2^5.

Therefore a = 11 and b = 5.

Duke_Of_Havoc