Hi sir. I solved this equation in a different way, splitting 2016 into its factors. Doing so , I got 2⁵.3².7 , which I further wrote as 2⁵.(2³+2⁰).(2³-2⁰) and then using the identity of (a+b)(a-b) I got 2ª - 2^b = 2¹¹ - 2⁵. Thanks a lot for the wonderful problems..helping us students a lot
Gostei desse método. Apenas vejo que seria por tentativas, que nesse caso já encontrou na primeira a diferença sendo uma potência de base 2. Mas se não encontrasse teria que tentar o segundo valor de "a", o terceiro, ... até encontrar como diferença uma potência de base 2.
I prime factored 2016 , and could see that 63 was not a power of 2. The substitution of a = b+k and resulting factoring enabled you to express an odd number. I had seen one of these problems before but could not remember how to express the odd factor. I will remember this in the future. Well done.
0:00 Let's try First power of 2 > 2016 is 2048 and 2048 - 2016 = 32 This gives the easy solution a=11 and b=5 Now for any n>=12, 2^n - 2^(n-1) = 2^(n-1) >= 2048 > 2016 This means that for any a >= 12, we have (a-1) < b < a so b cannot be a natural and our solution above is the only solution.
I think there is an easier way to go about this. 2^a has to be larger than 2016. So start listing powers of 2 that are larger than 2016. The first number that fulfills this is 2048 Then solve 2048 - x = 2016...x=32. 32 is also a power of 2, so we are done 2^a = 2048...a=11 2^b = 32...b=5
Brute force is the last resort. Your approach works in this case because the numbers are small. What if the difference was a 10 or 20 digit number instead of 2016? Then whose method-yours or the professor’s-is better?
That is almost exactly what I did Nate. I went a little bigger with 2 ^ 12 = 2 ^6 x 2^6 = 64 x 64 = 4096. 4096 - 2016 = 2080, and that is not a valid solution. I then divided 4096 by 2 to go with 2 ^ 11 to get 2048. 2048 - 2016 = 32. 2^5 = 32. Therefore, a = 11 and b =5.
Interesting! I solved it just the visual way. 2016 is binary 11111100000, so I just have to pick the next higher power of two, 100000000000, and subtract from it the smaller power of two, that's capable of just swapping 0s to 1s on the following 5 positions, 100000. So I only have to count the 5 1s (needed to swap back to from 0 in the new number by subtraction) and 11 0s in the new number. So 2¹¹ - 2⁵ => a=11, b=5 .
Doing the binary conversion seems like the hard way if you know your powers of two that well in the first place. Of course, just immediately recognizing the answer isn't something you can put down on paper to convince anyone you legitimately solved the problem.
Two shortcuts that make this problem a bit easier: The closest power of 2 that is greater than 2016 is 2048 or 2^11, and 2048 - 2016 = 32 = 2^5, so a = 11, b = 5. Alternatively, writing 2016 in binary and knowing how binary arithmetic works may provide a great hint: 2016 = 1,111,100,000 = 10,000,000,000 - 100,000 = 2^11 - 2^5.
Thank you! I enjoyed your solution. A more "nerdy" than mathematic approach comes as followed: 1. A power of 2 (2^n) has only one "1" in binary representation (a "1" and n-1 0's) 2. 2016 is "0111 1110 0000" . Add 1 to the last "1" in this number to get a power of 2. 3. "0111 1110 0000" + "0010 0000" = "1000 0000 0000", or 2016 + 32 = 2048. 4. a = log2(2048) = 11 and b = log2(32) = 5.
Wow!! Sir, you are like the Sherlock Holmes of mathematics. i love the way you gradually build up your evidence and then assemble it into a conclusion. This question is a bit advanced for me, because I would never have thought to approach this problem so systematically. Nevertheless, I thoroughly enjoy your explanations of how facts fit together, even if I get a bit lost sometimes. Thank you for posting your brilliant videos. It was through your videos that I learnt how to solve quadratic equations. I love doing them, especially the completing the square method. May God bless you in the Lord Jesus Christ.
The sum is really great, at the beginning I could understand nothing of the sum and was totally confused, but the way you explained is really appreciable.
Hello, Sir!. Love and respect from Italy. Since I'm particularly interested in these kind of exercises and all the stuff about math olympiads, would you mind suggesting me some resources to study the theory?. I struggle a bit with some questions, but I still adore you content. Thank you, sir. Keep going!
Hello dear, I don't know of any good book that I could recommend you. If I find any quality book, sure I'd love to share with you. Meanwhile, I'd recommend you to watch all the videos (on daily basis) in premath channel to get the deeper understanding. Thanks for asking. I wish you all the best. You are awesome. Keep it up 😀 Love and prayers from the USA!
I participated in the Austrian Mathematical Olympiad. The geometry teacher we had at our preparation course highly praised the book „Euclidean Geometry in Mathematical Olympiads“ by Evan Chen. I plan on getting it myself to practice for next year’s olympiad, but from what I’ve seen it is rather pricey at about 50€ on Amazon. Unfortunately, the other teachers didn’t mention any books and instead handed out self-written scripts only available to those who participate at the olympiad. Some of these scripts are available at the website of Austrian olympiad though, so you could check if the Italian olympiad offers similar studying materials or mentions any recommended books for studying.
@@Alexander-tx4bw The book you mentioned is quite expensive and apparently not really simple to find: it must be remarkable!. Thank you so much for your recommendation!.
Herr Professor, how are you? I could have thought the way you thought, which is simple, no need to do any math. (2048 - 32 ). But it didn't occur to me, what occurred to me was another solution that also worked ( 4096 - 2080) , where b = 11.02237. As it was the first time I saw such a question, my brain from now on will reason more simply and I will never be wrong again. Hugs from the South of Brazil.
In problems of this kind, I think that a good strategy is to first factorize the constant term so that 2016 = 32 * 63 = 2^5 * (64 -1 ) = 2^5 * (2^6 -1) from which we can write the values of a and b by inspection : a = 2^5 * 2*6 = 2^11 ; b = 2^5
Representing a number as the sum or difference of different powers of 2 is equivalent to converting to binary. It is a unique transformation. So look for the power of 2 that is next higher than 2016. It is 2048 ie 2^11 2048-2016= 32=2^5 Thus a=11 b=5
I'm deeply thankful for your content, it's a blessing from God that you exist to share your knowledge so that everyone who watches you can learn a little more each day. Cheers, mate!
Всё гораздо прозаичнее. Находим ближайшую степень 2, больше, чем 2016. Из уроков информатики помню, что 2^10 =1024, значит 2^11=2048. => a=11. 2048-2016=32=2^5. b=5 2^10 - 2^5 = 2016. Это решается в уме
Во первых в таком решении нет гарантии, что ближайшая степень двойки выше 2016 окажется первым корнем, а во-вторых нет доказательства, что подобранное решение единственно возможное и нет других действительных корней уравнения
One thing I have learnt for the first time that the multiplefixation of even and an odd number is equal to the corresponding numbers on the other side of the equation... Is there some rule accordingly....?
When he related (a) and (b) to the set of Naturals, it became easy, and since the bases were the same, I applied the trial and error method, as friends call "to apply the Bézout", even though Bezout has no application in the case, but it's how we usually work the issues that we need more attention...2¹⁰=1024, 2ª = 2¹¹=2048; and 2⁵=32...but your explanation cleared all doubts!!! a=b+k
10 seconds: you get 2016 if you substract 32 from 2048. 2048=2^11, 32=2^5. So a=11 and b=5. Since we were looking for natural numbers, I tried the most obvious solutions for powers of two first. (By the way: if you count binary, you can count to 1023 with 10 fingers… somehow, this never got a widespread thing, except maybe for early computer programmers…) 😀
I would love to learn binary approach. Could you please suggest any good resources on this please? Also how do you solve this problem using binary system? Please enlighten me.
@@mahtabalam7834 I do not have a source but it is actually quite simple: In our decimal system, each number is made up of digits that can take 10 different forms (0,1, 2,... 8, 9). If you take the nuber 321, it actually says "take 3 times 10³ plus 2 times 10² plus 1 time 10^0" which gives you 300+20+1=321. In the binary system you don't have 10^something, but 2^something and you only have the digit forms 0 and 1. (This is why it is used in IT, as a current can flow or not, on vs off.) The number 5 would be "101" in binary code, as it is 1*2^2+0*2^1+1*2^0 or 4+0+1=5. 31 would be "11111" or 1*2^4+1*2^3+1*2^2+1*2^1+1*2^0 or 16+8+4+2+1=31. This means that you can count to 31 with just one hand. 1 is our thumb extended. 10 is indexfinger extended, thumb retracted. 11 (which is 1+2=3) is both thumb and indefinger extended. And so on. Extending all 5 fingers of one hand is 11111 or 31. If you combine this with our second hand, you can count until 1023 as "1111111111" stands for 2^0+2^1+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9 or 1+2+4+8+16+32+64+128+256+512= 1023. (As you only have 10 fingers, you can only go up to 2^9, as you have to start with 2^0...)
Sir you could also have gone in this approach by using factors of 2016 2016=2^5×(63) So dividing both sides by 2^5 we get, (2^a-2^b)/2^5=63 2^(a-5)-2^(b-5)=63 Nearest multiple of 2 near to 63 is 64 or 2^6 -1 and 1 can be written as 2^0=1 So the equation becomes 2^{a-5}-2{b-5}=2^6-2^0 Comparing we get a-5=6 or a=11 And b-5=0 or b=5
For junior high school students and or their parents interested for the their children in learning advanced mathematics in junior classes, may watch :Olympiad Primer:th-cam.com/video/N6DOwMtq2Cc/w-d-xo.html
Completed by another method mentally in 2 minutes by substracting 2016 from nearest 2 power series number 2048, which is 2^11, then equating both sides terms individually which gives 2^11- 2^5 = 2048 - 32 = 2016 , so by equating this gives a= 11, b= 5
There's a very pretty solution utilising the uniqueness of binary representation 2^a = 2^b + 2016 2016 in binary is 11111100000 so we get that 2^a = 2^b + 2⁵ + 2⁶ + 2⁷ + 2⁸ + 2⁹ + 2¹⁰ Now we note by unqiness of binary representation that b € {5,6,7,8,9,10} Now suppose that b>5 Then the RHS = 2⁵ + k×2⁵ = (k+1)2⁵ where k is even so (k+1) is odd so the RHS contains at least 2 different powers of 2 which is again impossible by uniqueness of binary representation Hence b=5 and a=11
a > b a = 12 then 2^a - 2^b is more than or equal to 2048 > 2016 (minimum at b = 11 < a = 12) a = 10 then 2^a - 2^b < 1024 < 2016 Since a, b are positive integers, a = 11 Substitute in the original equation to get b = 5
Wow, similar to what I did, but more methodically rigorous. I got to: (2^b)[2^(a -- b) -- 1] = 2^5 × 3^2 × 7 The rest is easy. I was thinking I got lucky, mostly. 😊
At 4:29 you write Even*Odd = even*odd and then assert Even = even and Odd = odd. This does not follow.. For example 10*7 = 14*5. Your assertion requires stating that the even numbers are powers of 2.
Hi, amazing video once again! I asked this in an earlier video's comment section but IDK if you noticed: I love these videos but could it be possible to make these with a dark background instead of the white?
Thanks for the idea! I'll discuss with my technical team. Thanks for the visit Thank you for your feedback! Cheers! You are awesome, Eemeli. Keep it up 😀
I’d suggest there would be an infinite number of solutions since the constraints that a > b and a=b+k are too loose to produce a single solution. It looks like a variation, but a close relative of, the “Catalan conjecture”.
let me tell you my method of approaching first taking a greater than b if we not take then value will be negative so we have taken that then taking common 2^b (2^a-b -1 )= 2^5multiply 63 and now by equatin first with first term and 2 with 2 we et our ans.easy question
2016=0x7e0=0b 111,1110,0000. The right most 1 is index 5 (right most bit, or least significant bit, is index 0). Then b=5. The left most 1 is index 10, thus a = 10+1= 11.
First we determine that a > b, so we can transform the left side of equation into 2^b • (2^k-1) Next we note that expression 2^k is always even, and 2^k-1 is always odd. Then we must express right side of the equation into the product of the power of two by some number and some odd number. That can be done by subsequent dividing 2016 by two until the result is odd. So 2016 = 2^5 • 63 where b = 5 and 2^k-1 = 63 2^k = 64 = 2^6 k = 6 a = b+k = 11 Solved.
You could just as conveniently said 2^a greater than 2016, hence try next power of 2 : 2^11 = 2048 and so 2^b = 32 > b = 5 solves the problem. Just another way of making a guess. Although, it is true that 2^b must be a divisor of 2016, since 2^b is a divisor of 2^a.
if one observes that 2^11 = 2048 and 2^5 = 32 and subtract 32 from 2048 you'll can read the immediate answer for a=11 and b=5. Low level programers are familiar with the powers of 2.
2016=32*9*7 so the only way it could be of the form 2^x(2^y-1) is if x=5 because 2^5=32 and the part inside the bracket is odd. 9*7=63 so if we add 1 then we get 64 which is 2^6. which means 2016 = 2^5 (2^6-1)= 2^11 - 2^5
I solved this problem by adding 2^5 to both sides. 2^a - 2^b + 2^5 = 2016 + 2^5 = 2016 + 32 = 2048 = 2^11 So 2^a - 2^b = 2^11 - 2^5 So a=11 b=5 by comparison
I did it like this, I converted 2016 into its exponential factors i.e. 2¹¹-2⁵ and then just compared these values with 2^a - 2^b. so I compared the negative term with the negative and the positive one with the positive and my answers were a=11 and b=5. don't judge me pls, I'm just in 10th standard so I solved it with what I know till now
Since you have only one equation and 2 variables there should be an unlimited number of solutions. But you also know that a and b are natural positive numbers. So you can show that the difference between a and b is the lowest when a=b+1. and thus you won't find another solution/differences that match since 2^12-2^11=2048 and 2^10=1024. 2^a has to be =2^11
Since 2ᵃ must be a power of 2, I the first power of 2 greater than 2016. That is 2048 (2¹¹). 2ᵇ must also be a power of 2 that can be subtracted from 2048 to produce 2016. 2048 - 2016 = 32. 32 is 2⁵. Therefore a is 11 and b is 5.
My Procedure was different Consider next higher exponential values of 2, That is(2048), Considered it (2048) as 2^a Calculated difference of 2016 and 2048 (32), that is considered as 2^b It implied a=11 b=5
While I respect the manipulation used here, the fact is that the solution was really guessed rather than calculated. The reason for selecting 2^5 as a factor in the first place was that it was obvious that 5 was one of the anwers. I just guessed the answers directly, by "inspection" of the original equation, without any manipulation. I think that was simpler. But it's disappointing that there apparently is no method of calculating the solutions -- you just have to recognize them at some point, either with or without manipulation.
Dear JR, these are Olympiad questions! These questions are designed to check your manipulation techniques and confidence. I agree with you about the choice of numbers. Thank you for your feedback! Cheers! You are awesome. Keep it up 😀
2^a - 2^b = 2016 Seja a > b, tem-se que: 2⁵.(2^(a - 5) - 2^(b - 5)) = 2⁵.63 2^(a - 5) - 2^(b - 5) = 63 logo, o segundo termo há de ser ímpar e como é uma potência de base 2, então 2^(b - 5) = 1 => b - 5 = 0 => b = 5. E, consequentemente, 2^(a - 5) - 1 = 63 => 2^(a - 5) = 64 = 2⁶ => a - 5 = 6 => a = 11 (11, 5) é a única solução natural.
answer a=11 , b=5 2^a =2016 + 2^b hence 2^a is greater than 2016 , if assume 'a' and 'b' are integers then 2^a is at least 2048 or 2^11 , let assume it is 2^11 then 2^b = 2048-2016= 32 or 2^5. Since assume 'a' and 'b' are integers 'a'=11 and 'b'=5
I knew 2^11 was 2048. I subtracted 2016 from 2048. Bingo! The answer was 32 which can be expressed as 2^5 Now I wrote the problem as 2^a - 2^b = 2^11 - 2^5 So upon inspection I got a = 11 and b = 5
Observations: 1. 2^a must end with 8 and 2^b must end with 2 2. The positive powers of two - 2, 4, 8, 16, 32, 64, 128, 256... follows repeating pattern in their ending digit: 2, 4, 8, 6, 2, 4, 8, 6, 3. Last digit 8 begins the pattern h 2^3, 2,^7, 2^11 4. 2^11 = 2048 is close to required answer 2016 5. 2048-2016 =32 , hence b =5 and a^11
I think it can be done in a much more easier way. Look for a 2 power that is larger than 2048. 2^11 = 2048 2016 = 2048 - 32 2048 - 32 = 2^11 - 2^5 2^a-2^b = 2^11-2^5 a=11, b=5
How is this olympiad problem? For which class/ grade? It would be nice if you would have solve it in general case 2^a+2^b=c. Choosing right side as 2016 was like cheating . It solved the exercise by itself
You re' Awesome...Love from INDIA
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So nice of you, Rishit
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Hi sir. I solved this equation in a different way, splitting 2016 into its factors. Doing so , I got 2⁵.3².7 , which I further wrote as 2⁵.(2³+2⁰).(2³-2⁰) and then using the identity of (a+b)(a-b) I got 2ª - 2^b = 2¹¹ - 2⁵. Thanks a lot for the wonderful problems..helping us students a lot
Wow
You are realy clever
well done. Very nice. Clever.
Nice
This is great. But in this way, we still need one more thing: proving this is only one result.
2^a to be bigger than 2016.
First value of "a" to meet this requirement.
This was 2^11 = 2048.
2048 - 2016 = 32 = 2^5
The first power of 11 worked!
Thank you for your feedback! Cheers!
You are awesome, Monty. Keep it up 😀
Yes, first value of a came to mind is 11, I think you Solve it orally👍👍
Ya, I did the same. Did it in my head. I just wanted a feel for what the answer might be. Worked out.
Gostei desse método. Apenas vejo que seria por tentativas, que nesse caso já encontrou na primeira a diferença sendo uma potência de base 2. Mas se não encontrasse teria que tentar o segundo valor de "a", o terceiro, ... até encontrar como diferença uma potência de base 2.
I prime factored 2016 , and could see that 63 was not a power of 2. The substitution of a = b+k and resulting factoring enabled you to express an odd number. I had seen one of these problems before but could not remember how to express the odd factor. I will remember this in the future. Well done.
0:00 Let's try
First power of 2 > 2016 is 2048
and 2048 - 2016 = 32
This gives the easy solution a=11 and b=5
Now for any n>=12, 2^n - 2^(n-1) = 2^(n-1) >= 2048 > 2016
This means that for any a >= 12, we have (a-1) < b < a so b cannot be a natural and our solution above is the only solution.
I think there is an easier way to go about this.
2^a has to be larger than 2016. So start listing powers of 2 that are larger than 2016.
The first number that fulfills this is 2048
Then solve 2048 - x = 2016...x=32. 32 is also a power of 2, so we are done
2^a = 2048...a=11
2^b = 32...b=5
Many approaches possible.
Thank you for your feedback! Cheers!
You are awesome. Keep it up 😀
amazing ...th-cam.com/video/4v5sqzhaAlI/w-d-xo.html
That's how i've done it as well
Brute force is the last resort. Your approach works in this case because the numbers are small. What if the difference was a 10 or 20 digit number instead of 2016? Then whose method-yours or the professor’s-is better?
That is almost exactly what I did Nate. I went a little bigger with 2 ^ 12 = 2 ^6 x 2^6 = 64 x 64 = 4096. 4096 - 2016 = 2080, and that is not a valid solution. I then divided 4096 by 2 to go with 2 ^ 11 to get 2048. 2048 - 2016 = 32. 2^5 = 32. Therefore, a = 11 and b =5.
Interesting! I solved it just the visual way. 2016 is binary 11111100000, so I just have to pick the next higher power of two, 100000000000, and subtract from it the smaller power of two, that's capable of just swapping 0s to 1s on the following 5 positions, 100000. So I only have to count the 5 1s (needed to swap back to from 0 in the new number by subtraction) and 11 0s in the new number. So 2¹¹ - 2⁵ => a=11, b=5 .
Doing the binary conversion seems like the hard way if you know your powers of two that well in the first place. Of course, just immediately recognizing the answer isn't something you can put down on paper to convince anyone you legitimately solved the problem.
Nice method. I have an alternative solution.
Prime factorization of 2016 = 2^5 * 3^2 * 7
= 2^5 * (2^3 + 1) * (2^3 - 1)
= 2^5 * (2^6 - 1) [difference of squares]
= 2^11 - 2^5
a = 11, b = 5.
Hi Sir,
x^m- x^n= x^n and n=m-1
X=?
@@radhikar6690
X=2
Two shortcuts that make this problem a bit easier:
The closest power of 2 that is greater than 2016 is 2048 or 2^11, and 2048 - 2016 = 32 = 2^5, so a = 11, b = 5.
Alternatively, writing 2016 in binary and knowing how binary arithmetic works may provide a great hint: 2016 = 1,111,100,000 = 10,000,000,000 - 100,000 = 2^11 - 2^5.
Thank you! I enjoyed your solution.
A more "nerdy" than mathematic approach comes as followed:
1. A power of 2 (2^n) has only one "1" in binary representation (a "1" and n-1 0's)
2. 2016 is "0111 1110 0000" . Add 1 to the last "1" in this number to get a power of 2.
3. "0111 1110 0000" + "0010 0000" = "1000 0000 0000", or 2016 + 32 = 2048.
4. a = log2(2048) = 11 and b = log2(32) = 5.
Excellent! Nice approach..
Thank you for your feedback! Cheers!
You are awesome. Keep it up 😀
Hi Sir,
x^m- x^n= x^n and n= m-1
X=?
Wow!! Sir, you are like the Sherlock Holmes of mathematics. i love the way you gradually build up your evidence and then assemble it into a conclusion. This question is a bit advanced for me, because I would never have thought to approach this problem so systematically. Nevertheless, I thoroughly enjoy your explanations of how facts fit together, even if I get a bit lost sometimes. Thank you for posting your brilliant videos. It was through your videos that I learnt how to solve quadratic equations. I love doing them, especially the completing the square method. May God bless you in the Lord Jesus Christ.
You're welcome!
Glad you enjoyed it!
Thank you for your feedback! Cheers!
You are awesome, William. Stay blessed 😀
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amazing ...th-cam.com/video/4v5sqzhaAlI/w-d-xo.html
an easier way is to replace 2016 by 2048-32, means 2^11 - 2^5
We have then to solve
2^a - 2^b = 2^11 - 2^5
by identifying , we can find
a=11
b=5
The sum is really great, at the beginning I could understand nothing of the sum and was totally confused, but the way you explained is really appreciable.
Hello, Sir!. Love and respect from Italy. Since I'm particularly interested in these kind of exercises and all the stuff about math olympiads, would you mind suggesting me some resources to study the theory?. I struggle a bit with some questions, but I still adore you content. Thank you, sir. Keep going!
Hello dear, I don't know of any good book that I could recommend you. If I find any quality book, sure I'd love to share with you. Meanwhile, I'd recommend you to watch all the videos (on daily basis) in premath channel to get the deeper understanding. Thanks for asking. I wish you all the best.
You are awesome. Keep it up 😀
Love and prayers from the USA!
I participated in the Austrian Mathematical Olympiad. The geometry teacher we had at our preparation course highly praised the book „Euclidean Geometry in Mathematical Olympiads“ by Evan Chen. I plan on getting it myself to practice for next year’s olympiad, but from what I’ve seen it is rather pricey at about 50€ on Amazon. Unfortunately, the other teachers didn’t mention any books and instead handed out self-written scripts only available to those who participate at the olympiad. Some of these scripts are available at the website of Austrian olympiad though, so you could check if the Italian olympiad offers similar studying materials or mentions any recommended books for studying.
@@Alexander-tx4bw The book you mentioned is quite expensive and apparently not really simple to find: it must be remarkable!. Thank you so much for your recommendation!.
Herr Professor, how are you?
I could have thought the way you thought, which is simple, no need to do any math. (2048 - 32 ). But it didn't occur to me, what occurred to me was another solution that also worked ( 4096 - 2080) , where b = 11.02237. As it was the first time I saw such a question, my brain from now on will reason more simply and I will never be wrong again. Hugs from the South of Brazil.
In problems of this kind, I think that a good strategy is to first factorize the constant term so that
2016 = 32 * 63 = 2^5 * (64 -1 ) = 2^5 * (2^6 -1)
from which we can write the values of a and b by inspection : a = 2^5 * 2*6 = 2^11 ; b = 2^5
And you must prove this is only one result.
@@buileminhquoc1675 a and b are natural number as a given Also radical solutions are excluded since 2016 in an integer..
Representing a number as the sum or difference of different powers of 2 is equivalent to converting to binary. It is a unique transformation.
So look for the power of 2 that is next higher than 2016.
It is 2048 ie 2^11
2048-2016= 32=2^5
Thus a=11 b=5
I'm deeply thankful for your content, it's a blessing from God that you exist to share your knowledge so that everyone who watches you can learn a little more each day. Cheers, mate!
You're welcome!
Thank you for your feedback! Cheers!
You are awesome, Lucas. Keep it up 😀
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amazing ...th-cam.com/video/4v5sqzhaAlI/w-d-xo.html
Hello, Sir! Love and respect from Russia. We adore you video. Thank you, sir, for your content. Keep going!
Всё гораздо прозаичнее. Находим ближайшую степень 2, больше, чем 2016. Из уроков информатики помню, что 2^10 =1024, значит 2^11=2048. => a=11. 2048-2016=32=2^5. b=5
2^10 - 2^5 = 2016. Это решается в уме
Во первых в таком решении нет гарантии, что ближайшая степень двойки выше 2016 окажется первым корнем, а во-вторых нет доказательства, что подобранное решение единственно возможное и нет других действительных корней уравнения
One thing I have learnt for the first time that the multiplefixation of even and an odd number is equal to the corresponding numbers on the other side of the equation...
Is there some rule accordingly....?
When he related (a) and (b) to the set of Naturals, it became easy, and since the bases were the same, I applied the trial and error method, as friends call "to apply the Bézout", even though Bezout has no application in the case, but it's how we usually work the issues that we need more attention...2¹⁰=1024, 2ª = 2¹¹=2048; and 2⁵=32...but your explanation cleared all doubts!!! a=b+k
10 seconds: you get 2016 if you substract 32 from 2048. 2048=2^11, 32=2^5. So a=11 and b=5. Since we were looking for natural numbers, I tried the most obvious solutions for powers of two first. (By the way: if you count binary, you can count to 1023 with 10 fingers… somehow, this never got a widespread thing, except maybe for early computer programmers…) 😀
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I would love to learn binary approach. Could you please suggest any good resources on this please? Also how do you solve this problem using binary system? Please enlighten me.
@@mahtabalam7834 I do not have a source but it is actually quite simple: In our decimal system, each number is made up of digits that can take 10 different forms (0,1, 2,... 8, 9). If you take the nuber 321, it actually says "take 3 times 10³ plus 2 times 10² plus 1 time 10^0" which gives you 300+20+1=321. In the binary system you don't have 10^something, but 2^something and you only have the digit forms 0 and 1. (This is why it is used in IT, as a current can flow or not, on vs off.) The number 5 would be "101" in binary code, as it is 1*2^2+0*2^1+1*2^0 or 4+0+1=5. 31 would be "11111" or 1*2^4+1*2^3+1*2^2+1*2^1+1*2^0 or 16+8+4+2+1=31. This means that you can count to 31 with just one hand. 1 is our thumb extended. 10 is indexfinger extended, thumb retracted. 11 (which is 1+2=3) is both thumb and indefinger extended. And so on. Extending all 5 fingers of one hand is 11111 or 31. If you combine this with our second hand, you can count until 1023 as "1111111111" stands for 2^0+2^1+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9 or 1+2+4+8+16+32+64+128+256+512= 1023. (As you only have 10 fingers, you can only go up to 2^9, as you have to start with 2^0...)
@@philipkudrna5643 wow! It's awesome. Will try it now. Thank you so much sir.
Sir, awasom an easier method for a difficult problem,you are great.
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Thank you for this question
Knowing that 2 to the power of 10 is 1024 , 2 to the 11 is 2048. 48-16 = 32 = 2 to p of 6
A= 11 B=6
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Unless that 32=2^5 :-p
That's how bugs in programs happen. ;-)
Sir you could also have gone in this approach by using factors of 2016
2016=2^5×(63)
So dividing both sides by 2^5 we get,
(2^a-2^b)/2^5=63
2^(a-5)-2^(b-5)=63
Nearest multiple of 2 near to 63 is 64 or 2^6 -1 and 1 can be written as 2^0=1
So the equation becomes
2^{a-5}-2{b-5}=2^6-2^0
Comparing we get
a-5=6 or a=11
And b-5=0 or b=5
Great, this is what I was getting ready to nsuggest.
I mean suggest
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Hi Sir,
x^m- x^n= x^n and n=m-1
X=?
For junior high school students and or their parents interested for the their children in learning advanced mathematics in junior classes, may watch :Olympiad Primer:th-cam.com/video/N6DOwMtq2Cc/w-d-xo.html
Watching your videos on regular basis made me solve even tough problems 💝😃
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Completed by another method mentally in 2 minutes by substracting 2016 from nearest 2 power series number 2048, which is 2^11, then equating both sides terms individually which gives 2^11- 2^5 = 2048 - 32 = 2016 , so by equating this gives a= 11, b= 5
There's a very pretty solution utilising the uniqueness of binary representation
2^a = 2^b + 2016
2016 in binary is 11111100000 so we get that
2^a = 2^b + 2⁵ + 2⁶ + 2⁷ + 2⁸ + 2⁹ + 2¹⁰
Now we note by unqiness of binary representation that b € {5,6,7,8,9,10}
Now suppose that b>5
Then the RHS = 2⁵ + k×2⁵ = (k+1)2⁵ where k is even so (k+1) is odd so the RHS contains at least 2 different powers of 2 which is again impossible by uniqueness of binary representation
Hence b=5 and a=11
a > b
a = 12 then 2^a - 2^b is more than or equal to 2048 > 2016 (minimum at b = 11 < a = 12)
a = 10 then 2^a - 2^b < 1024 < 2016
Since a, b are positive integers, a = 11
Substitute in the original equation to get b = 5
So logical. Spock-like.
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Awesome. What a trick. Respect you sir.
Wow, similar to what I did, but more methodically rigorous. I got to:
(2^b)[2^(a -- b) -- 1] = 2^5 × 3^2 × 7
The rest is easy.
I was thinking I got lucky, mostly. 😊
At 4:29 you write Even*Odd = even*odd and then assert Even = even and Odd = odd. This does not follow.. For example 10*7 = 14*5. Your assertion requires stating that the even numbers are powers of 2.
Excellent! Awesome!!
Saraha ma3andich zhar la f tssahib la f zwaj walit tangoul tawahad maynawad tawahda Matahmal
Nice technique on the (32)(63) factoring of 2016
Great
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You are awesome sir
You are really awesome. Thanks a lot
Hi, amazing video once again! I asked this in an earlier video's comment section but IDK if you noticed: I love these videos but could it be possible to make these with a dark background instead of the white?
Thanks for the idea! I'll discuss with my technical team.
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It's a brilliant, Sherlock !
Thanks for video. Good luck sir!!!!!!!!
Very nice!!!!!
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Can you prove that these are the only values a,b satisfying the constraints?
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I’d suggest there would be an infinite number of solutions since the constraints that a > b and a=b+k are too loose to produce a single solution.
It looks like a variation, but a close relative of, the “Catalan conjecture”.
Yup. Here is the complete answer:
2^a !=2016 and b = (log(2^a - 2016) + 2 i π n)/log(2) and n element Z
Wolfram Alpha
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Excellent. Thank you.
I also used your method, ty!
Very nice solution. Excellent presentation.
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love the question bro, you explained the solution very well, great job
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Very good!!!
ok , I understand and I try to know and learn more .
let me tell you my method of approaching first taking a greater than b if we not take then value will be negative so we have taken that then taking common 2^b (2^a-b -1 )= 2^5multiply 63 and now by equatin first with first term and 2 with 2 we et our ans.easy question
Hi Sir,
x^m- x^n= x^n and n=m-1
X=?
@@radhikar6690 x will be 2 i think let me know or i will solve with another method
@@radhikar6690 why you say me sir 🤣🤣i am a idiot student in my class i just use my aptitude of class7 to solve this question
@@radhikar6690 tell me
2048-32 popped into my head instantly. 2^5=32 I could figure out in my head. For 2^11=2048 I needed all my fingers… and one toe :)
2016=0x7e0=0b 111,1110,0000. The right most 1 is index 5 (right most bit, or least significant bit, is index 0). Then b=5. The left most 1 is index 10, thus a = 10+1= 11.
Another awesome video👍
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Wow! Well done 👍🏻
always great video !
At 4:30 how do you proove unicity ?
Lovely method.
First we determine that a > b, so we can transform the left side of equation into
2^b • (2^k-1)
Next we note that expression 2^k is always even, and 2^k-1 is always odd.
Then we must express right side of the equation into the product of the power of two by some number and some odd number. That can be done by subsequent dividing 2016 by two until the result is odd.
So 2016 = 2^5 • 63
where b = 5 and
2^k-1 = 63
2^k = 64 = 2^6
k = 6
a = b+k = 11
Solved.
You could just as conveniently said 2^a greater than 2016, hence try next power of 2 : 2^11 = 2048 and so 2^b = 32 > b = 5 solves the problem.
Just another way of making a guess.
Although, it is true that 2^b must be a divisor of 2016, since 2^b is a divisor of 2^a.
Fantastic working.👍👍
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if one observes that 2^11 = 2048 and 2^5 = 32 and subtract 32 from 2048 you'll can read the immediate answer for a=11 and b=5. Low level programers are familiar with the powers of 2.
2016=32*9*7 so the only way it could be of the form 2^x(2^y-1) is if x=5 because 2^5=32 and the part inside the bracket is odd. 9*7=63 so if we add 1 then we get 64 which is 2^6. which means 2016 = 2^5 (2^6-1)= 2^11 - 2^5
Can’t I factorize 2016 in other ways than 32*63? So the results can be different as long as a>b ?
Hello. What note-taking app do u use?
From the perspective of a software engineer, this problem is super easy no brainer...
Excellent sir
Superb! !!!
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Or :
2^(b+k) + 2^b = 2016
(2^b)(2^k) + 2^b = 2^11 + 2^5
b+k = 11 , while b = 5
Then ,
5+k = 11
k = 6
Therefore :
k = 6 and b = 5
Sir can you prove the following equation in the next video?
(1-x)-½=1+1/2x+3/8x²+5/16x³ ............................................................
Vvvvv nice explanation
So nice of you
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Hi! excellent work! What about a=10 and b=3?
That's not possible.
That would be less than 2016
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I solved this problem by adding 2^5 to both sides.
2^a - 2^b + 2^5 = 2016 + 2^5 = 2016 + 32 = 2048 = 2^11
So 2^a - 2^b = 2^11 - 2^5
So a=11 b=5 by comparison
I did it like this,
I converted 2016 into its exponential factors i.e. 2¹¹-2⁵ and then just compared these values with 2^a - 2^b. so I compared the negative term with the negative and the positive one with the positive and my answers were a=11 and b=5.
don't judge me pls, I'm just in 10th standard so I solved it with what I know till now
That is indeed an answer. But is it also proof that it is the only answer?
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Since you have only one equation and 2 variables there should be an unlimited number of solutions.
But you also know that a and b are natural positive numbers. So you can show that the difference between a and b is the lowest when a=b+1. and thus you won't find another solution/differences that match since 2^12-2^11=2048 and 2^10=1024.
2^a has to be =2^11
@@PeterLE2 Exactly! My point was that the given proof is not complete. If you add your explanation, it is.
nice solution
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I'm running out of superlatives to describe your questions and solutions!
You are too generous, my dear friend.
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Sir you are amazing
so cool ❤❤
THANK U ❤❤
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Since 2ᵃ must be a power of 2, I the first power of 2 greater than 2016. That is 2048 (2¹¹). 2ᵇ must also be a power of 2 that can be subtracted from 2048 to produce 2016. 2048 - 2016 = 32. 32 is 2⁵. Therefore a is 11 and b is 5.
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Hi Sir,
x^m- x^n= x^n and n=m-1
X=?
My Procedure was different
Consider next higher exponential values of 2, That is(2048), Considered it (2048) as 2^a
Calculated difference of 2016 and 2048 (32), that is considered as 2^b
It implied a=11
b=5
Sir can you solve for x in 2^x=5
I just splitted 2016 into 2048-32 and then 2¹¹-2⁵
Nice one
While I respect the manipulation used here, the fact is that the solution was really guessed rather than calculated. The reason for selecting 2^5 as a factor in the first place was that it was obvious that 5 was one of the anwers. I just guessed the answers directly, by "inspection" of the original equation, without any manipulation. I think that was simpler. But it's disappointing that there apparently is no method of calculating the solutions -- you just have to recognize them at some point, either with or without manipulation.
Dear JR, these are Olympiad questions! These questions are designed to check your manipulation techniques and confidence. I agree with you about the choice of numbers. Thank you for your feedback! Cheers!
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Muito bom!!!
Love from INDIA.
Thanks
2^a - 2^b = 2016
Seja a > b, tem-se que:
2⁵.(2^(a - 5) - 2^(b - 5)) = 2⁵.63
2^(a - 5) - 2^(b - 5) = 63
logo, o segundo termo há de ser ímpar e como é uma potência de base 2, então 2^(b - 5) = 1 => b - 5 = 0 => b = 5.
E, consequentemente, 2^(a - 5) - 1 = 63 => 2^(a - 5) = 64 = 2⁶ => a - 5 = 6 => a = 11
(11, 5) é a única solução natural.
We can write 216 as 2^11-2^5
And can compare a with 11 and b with 5..
answer a=11 , b=5
2^a =2016 + 2^b hence 2^a is greater than 2016 , if assume 'a' and 'b' are integers then
2^a is at least 2048 or 2^11 , let assume it is 2^11 then 2^b = 2048-2016= 32 or 2^5. Since assume 'a' and 'b'
are integers 'a'=11 and 'b'=5
Super
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thank you sir 😁
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please call me yoonho
@@SuperYoonHo Thanks Yoonho. I believe it's South Korean... You are the best 😀
@@PreMath you too i am south korean 이호석 is my father's name
i'm 이윤호
I knew 2^11 was 2048.
I subtracted 2016 from 2048. Bingo! The answer was 32 which can be expressed as 2^5
Now I wrote the problem as 2^a - 2^b = 2^11 - 2^5
So upon inspection I got a = 11 and b = 5
Observations:
1. 2^a must end with 8 and 2^b must end with 2
2. The positive powers of two - 2, 4, 8, 16, 32, 64, 128, 256... follows repeating pattern
in their ending digit: 2, 4, 8, 6, 2, 4, 8, 6,
3. Last digit 8 begins the pattern h 2^3, 2,^7, 2^11
4. 2^11 = 2048 is close to required answer 2016
5. 2048-2016 =32 , hence b =5 and a^11
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Okay but how do ya know that you gotta have 32 × 63 over there ? Like it could be anything else
How do I find out that those two are the ones I need ?
I think it can be done in a much more easier way.
Look for a 2 power that is larger than 2048.
2^11 = 2048
2016 = 2048 - 32
2048 - 32 = 2^11 - 2^5
2^a-2^b = 2^11-2^5
a=11, b=5
Work in IT? :o)
When I saw the question, the first move I made was 2048-32=2016, from here I understood that a=11 and b=5
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Hi Sir,
x^m- x^n= x^n and n=m-1
X=?
How is this olympiad problem? For which class/ grade? It would be nice if you would have solve it in general case 2^a+2^b=c. Choosing right side as 2016 was like cheating . It solved the exercise by itself
Sir , can we use logs as well?
Many approaches possible!
@@PreMath thanks for the reply, when I saw thi s at first logs seemed as first option because we have same bases in lhs
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