Another way to solve can be to see the binary representation of 328 which is 101001000 and converting it to power of 10s. It will be 10^3 + 10^6 + 10^8. Now replacing 10 with 2, it will become 2^3 + 2^6 + 2^8 which can be rewritten as 2^3 + 8^2 + 4^4 or 8^1 + 2^6 + 4^4 or 8^1 + 4^3 + 2^8

@dimondais2034

Жыл бұрын

Thanks..it's a good way

@gytoser801

Жыл бұрын

Wow

@vedantvyas3639

Жыл бұрын

👏

@punyanshjuneja941

Жыл бұрын

Fantastic

@ale4462

Жыл бұрын

What’s binary

Interesting. I see this as a combinatorics problem. Convert 328 to binary = 256 + 64 + 8 = 2^8 + 2^6 + 2^3. First choose 64 as 2^6, 4^3 or 8^2. 256 is not a power of 8. 8 is not a power of 4. Thus there is only one way to choose for 8 and 256. Hence the powers of 2,4,8 are (6,4,1) (8,3,1) (3,4,2).

@italixgaming915

Жыл бұрын

That's also my solution but don't forget to say that from the definition of binary system, there is only one way to write 328 as a sum of powers of 2.

@kenhaley4

Жыл бұрын

Yes, I like that approach much better than the video

@Bunny99s

Жыл бұрын

@@italixgaming915 That is true, but only because the binary system only allows each power to appear once. The variables a, b and c do not have this limitation. Also keep in mind that we only get a limited set of solutions because we restrict ourselfs to integers for a, b and c. If real numbers are allowed, we would have an underdetermined system since we have 3 unknowns and just one equation. Just the fact that we restrict the solutions to integers will give us a finite set of solutions. Note that if you have a power of two several times, say 3*2^4, you can always split the "3" into 2 and 1. The two becomes a 1 as you use the next higher power (so you simply carry over). So you get 2^5 + 2^4 == 3*2^4. This can also be seen as a binary bit shift. Multiplying by a power of two means you shift the places over by the exponent. So 3*2^4 is essentially the 3 represented as a binary 11b and then shifted 4 places to the left 110000b. This also gives us the 2^5 and 2^4

@yuvraj3941

Жыл бұрын

I found 3,4,2in my mind

@AT-rw3ou

Жыл бұрын

@@Bunny99s Pick any base, it’s representation will allow each power to appear exactly once, and I am not sure you can find a solution where a, b, c aren’t all non-negative integers.

Brute force is easier than the solution shown, IMHO C can only be 1 or 2 since 8^3>328, so the third term is either 8 or 64 If C=1, then the first two terms have to sum to 320. The only powers of two that sum to 320 are 256 + 64, which is either 2^8 + 4^3 or 2^6 + 4^4. So the first two solutions are 8,3,1 and 6,4,1 If C=2, then the first two terms have to sum to 264. The only powers of two that sum to 264 are 8+256, which can only be 2^3 + 4^4. That gives us the third solution 3,4,2. Done.

@Alex-ed8vj

9 ай бұрын

That's exactly how I did it and it is easier and shorter and easier to understand for students.

I like the way you crossed out the association between p, q, r and a, b, c. I didn’t do that step and it still works but you have to break it into cases which can be more work

Highly informative 👌

8, 3, 1, respectively. Get 2 raised to a value that is as close as possible to 328. 2^8 = 256. 328 - 256 = 72. 72 = 64 + 8. The two remaining exponents are self evident. Fun little exercise. Beats playing video games.

8^3 = 512 > 328, therefore c = 0 or 1 or 2. + c = 0: we have 2^a + 4^b = 327. Easily see that if both a >= 1 and b >= 1 then 2^a and 4^b are all evens, their sum will be an even. So a = 0 (leading to 4^b = 326, no b found) or b = 0 (leading to 2^a = 326, no a found). + c = 1: we have 2^a + 4^b = 320. 4^5 = 1024 > 320, therefore b = 0 or 1 or 2 or 3 or 4. - b = 0: 2^a = 319, no a found - b = 1: 2^a = 316, no a found - b = 2: 2^a = 304, no a found - b = 3: 2^a = 256, a = 8 - b = 4: 2^a = 64, a = 6 + c = 2: we have 2^a + 4^b = 264, b = 0 or 1 or 2 or 3 or 4. Similar technique with case c = 1, we find b = 4 and a = 3 So 3 pairs: (a,b,c) = (3, 4, 2); (6, 4, 1); (8, 3, 1) The general idea is to look at the variable with the biggest base and limit its potential.

Another easy brute force method: Rephrase the question: How many ways can you write 328 as the sum of a power of 2, a power of 4 and a power of 8? Start with the simpler question: which powers of two add up to 328? For reference, list the powers of two. 328 lies between 256 and 512, so one of the terms is 256. What's the next term? 328 minus 256 is 72, which is between 64 and 128, so one of the terms is 64. Subtract 64 from 72, which leaves 8 as the third term. We now know the three terms are 8, 64 and 256, in some order. So how can they be written as a power of 2, a power of 4 or a power of 8? Well, A power of 2 could be 8, 64 or 256, A power of 4 could be 64 or 256 (but not 8), A power of 8 could be 8 or 64 (but not 256). So what are the possible combinations in the form 2^a+4^b+8^c? If the first term (i.e. the power of 2) = 8, then the only way to complete the sum is 8+256+64 If the first term is 64, then the only way to complete the sum is 64+256+8 If the first term is 256, then the only way to complete the sum is 256+64+8 So there are exactly three distinct ways to write 328 as the sum of powers of 2, 4 and 8.

Very good ..nice learn mathemathic

Simply: The first term can be: 2, 4, 8, 16, 32, 64, 128 or 256. The second: 4, 16, 64 or 256. The third: 8 or 64. To get to 328, a 256 is essential, then it is clear that the other two numbers are 64 and 8. It is easy to pick out the possibilities. The method shown here is ingenious, but comparatively long winded and frankly not really necessary.

@ayushwithasingle_a

Жыл бұрын

The idea is not to get the solution but to get all the possible solutions and also to show that other solutions don't exists.

@claykarmel7720

Жыл бұрын

Martin is right, except he didn’t rule out 0 as an exponent. a, b or c can’t be zero because the left side would be odd. Two of them can’t be zero because 326 is not an integer power of 2, 4 or 8. A,b,c

Not bad, BUT there's an easier and faster way: 8^3 = 512 > 328 and 8^0 = 1 (can't be cuz the left side has to be even) so c can only be 1 or 2. For c = 1 => a = 6 and b = 4 or a = 8 and b = 3. For c = 2 => a = 3 and b = 4

@themangobui1474

Жыл бұрын

ayo I also solved it that way

@nitish18tayal

Жыл бұрын

The reason given for 8^0 is not valid as there are other ways for becoming even as if both 8 and 4 have 0 powers. In this case it's not possible because 328 is not a power of 2 but in general you can get an even number even if 8 has a power 0. Also you have not given reason for why a and b will have those specific values...

@danmimis4576

Жыл бұрын

@@nitish18tayal Indeed, that was so obvious that I just didn't insist on it, it was c < 3 that was important.

@nitish18tayal

Жыл бұрын

@@danmimis4576 Getting to c

@danmimis4576

Жыл бұрын

@@nitish18tayal A 5th grader can see that 326 (which is 328 - c^0 - a^0 or 328 - c^0 - b^0) is not a power of 2, NOT 328. What you said has 0 (ZERO) value: "In this case it's not possible because 328 is not a power of 2"

You can simplify by 8 before the replacement. Then it is easier as one p q r must be 0.

Thanks for video, verrygood

328=256+64+8= 2^8 + 2^6+2^3 implies the factors can not be summed up as a power of 2. 2^3 can be formed only by 2 or 8. C=1, a=8, b=3 C=1, a=6, b=4 A=3, c=2, b=4

For this specific question, a much simplified solution is that since 8^3=512>328, so c=1 or 2. When c=1 or 2, 2^a+4^b=320 or 264, then either can be solved by a same method but we only have 2 variables to deal with instead of 3 so it's much simpler.

@JohnSmith-nx7zj

Жыл бұрын

Yeah this doesn’t feel very hard to brute force. With c = 1 or 2, b

@007svikrant

Жыл бұрын

This way you can do it under 1 minute

@007svikrant

Жыл бұрын

I did it in about 50-60 secs

@007svikrant

Жыл бұрын

a=3, b=4 & c=2

I understood 100%. Make the mathematical structures at both side of the equations isomorphic to find the values of a, b and c.

Prime factoring 328 and taking common powers in Lhs simultaneously would make hit and trial more easier

It's a very simple method by putting a value.c is alaways less than 3 ,bcz (512) So we are putting two conditions c=1 or 2 By this we can also find a,b by possible values

very interesting approach

Trial and error does it much faster, although learning a different way doesn't hurt

It can be easily solve by two methods First method is convert 328 into binary Second method by trial and error method

Oh my gosh! This guy’s math is absolutely extraordinary

I have solved this from last digits concept Number. Last digit 2 2 2^2. 4 2^3. 8 2^4. 6(then it will repeat 4. 4 4^2 6. 8. 8 8^2. 4 8^3. 2 8^4. 6. After this I just calculate the possible ways And well I really like to see multiple approaches ti solve a problem

i agree with the methods aritten about comverting to binary. two ways to do it. repested division by two and the list of remainders in reverse order is the binary representation. or repeatedly subtract the largest power if two. ie. 256,64,8 2^8. 2^6,2^3 8^c cannot be in first position power not div. 3 4^b cannot be in last position power not div. 2 leaving 3 possible ordered *_triples_ full answer.. see other comments

the step at around 3:15 is technically not rigorous enough you are just assuming that 2^p=2^3 and (1+2^(q-p)+2^(r-p))=41. in theory it is also possible that e g 2^p=2^2 and (1+2^(q-p)+2^(r-p))=82. it is easy to see that the latter is in fact not possible (since that would require 2^(q-p)=1 for the lhs to be even and thus 80 would have to be a power of 2, which is not true), but it still wrong to completely neglect this possibility. if we had a 33 instead of 41 (which is also an odd number), then p=q=2 and r=8 would be a solution.

@maratibragimov1994

9 ай бұрын

Thank you for this. Had to scroll pretty bad to find this though. This comment should be on top and addressed by the author

328 can be written uniquely as a sum of powers of 2 as 328=2^8+2^6+2^3 Hence (a,2b,3c)=(8,6,3) then (a,b,c)=(8,3,1), (6,4,1) ,(3,4,2) .....

@irrelevant_noob

Жыл бұрын

That first statement is wrong if you don't restrict the number of terms within the sum... 328 can be written as a sum of powers of 2 also as 2^7+2^7+2^6+2^3, or 2^8+2^5+2^5+2^2+2^2, etc...

@DilipKumar-ns2kl

Жыл бұрын

@@irrelevant_noob It is obvious that 328 can be expressed as sum of powers of 2 uniquely involving only 3 terms as the problem suggests 328=2^8+2^6+2^3 Hence (a,2b,3c)=(8,6,3) It indicates that c is always equal to 1. Only we have to find (a,2b)=(8,6). If a=8, b=3 and if a=6, b=4 Hence solution sets (a,b,c) are (8,3,1) and (6,4,1).

@irrelevant_noob

Жыл бұрын

@@DilipKumar-ns2kl well, thank you for adding that "involving only 3 terms" i guess... which is exactly why i suggested to "restrict the number of terms". :-) But no, c is not "always equal to 1", since c=2, b=4, a=3 also works. -.- (You probably forgot about it, after these months... the OP had included it.)

Math is never about the answer but the way you approach it

@007svikrant

Жыл бұрын

Its about the answer if you have to aolve it in 40secs for competitive exam.

@dpage446

Жыл бұрын

@@007svikrant true, but you will never realize the beauty of maths if you only look at it from a competitive exam point of view.

@alphabeta2589

Жыл бұрын

@@007svikrant It's from Junior Balkan Math Olympiad. Subjective exam. You need to show an appropriate method else answer cancelled

He is finding the values by using isomorphism in the mathematical structure.

GREAT RESOLUTION!!

Why we take p

Excelente

To find ONE solution: LHS = 2^a + 4^b + 8^3; RHS = 328. The largest possible power of 2 is 256 = 2^8 which leaves 72. Given that, the largest possible power of 4 is 64 = 4^3 which leaves 8 Given those, the exponent of 8 must be 1: 8^1 = 8. So 2^8 + 4^3 + 8^1 = 256 + 64 + 8 = 328. a =8, b = 3, c = 1. Find the other combinations by "brute force," i.e. trial & error. Probably do-able, but I haven't done it.

Loved the solution and loved your voice.

@l.w.paradis2108

Жыл бұрын

Me too! You beat me to it. I was about to say the very same thing. :)

I actually solved it, but unfortunately, I got a rational number c, not a natural number 😅 2^a + 4^b + 8^c = 328 2^a + 2^(2b) + 2^(3c) = 328 2^a[1 + 2^(2b-a) + 2^(3c-a)] = 2³(41) 2^a = 2³ a = 3 1 + 2^(2b-a) + 2^(3c-a) = 41 2^(2b-a) + 2^(3c-a) = 40 2^(2b-a)[1 + 2^(3c-2b)] = 2³(5) 2^(2b-a) = 2³ 2b - a = 3 2b - 3 = 3 2b = 6 b = 3 1 + 2^(3c-2b) = 5 2^(3c-2b) = 4 2^(3c-2b) = 2² 3c - 2b = 2 3c - 2(3) = 2 3c - 6 = 2 3c = 8 c = 8/3 2^a + 4^b + 8^c = 328 2³ + 4³ + 8^(8/3) = 328 2³ + 4³ + 2^8 = 328 8 + 64 + 256 = 328

Using a bit of logic here. You start with 8. 8 cubed is to big so pick 8 squared. 328-64 remains. Now pick 4. 4 to the 5th it is to big. try 4 to the 4th. now we are left with an 8 which we all know is 2 to the third.

a,b,c are 3,4,2, respectively 2 to the power of 3 is 8 4 to the power of 4 is 256 and 8 square is 64 Total 328

Здравствуйте. Я не профессионал, а просто люблю решать различные задачи, и мне не понятно: 1. зачем решать через подстановки? можно сразу решать через а,в,с и получить конкретный ответ, что , кстати, и требуется в задаче. 2. почему: p《q 《j ? А вообще канал -- высший класс. Это первая задача, на которую я случайно наткнулась, и сразу подписалась.

The solution is WRONG until you consider q=p. In this case what's inside brackets becomes even and p CAN'T be 3, only 2 or 1. It doesn't give any new answer, but without considering it your solution is not full.

As already stated by others the solution is way too complicated. 2^a + 4^b + 8^c = 2^a + 2^(2b) + 2^(3c) = 328 = 2^8 + 2^6 + 2^3. 3c can only be 3 or 6 but not 8. in the case of 3c=6 only a can be 3 but not 2b so there are three solutions (3, 4, 2), (6, 4, 1) and (8, 3, 1).

V nice

Isme esa b to soch sakte h ke 8 me power 2 se jayda nhi jayege, or bacha 4 to uske power padate rho jitna vo total ke pas jaye usme 8 ka square jod do baki jitna no kam rha utna 2 ke power se add kr le ,mana ye koi method ni h per sayd esa b socha ja sakta h

Sehr nette Aufgabe. 👍

nice

if a = 0 and b = 0 max value x can take will be{1,2} So 8^{1,2} => {8,64} if a and c are 0 max value b can take will be {1,2,3,4} So, 4^{1,2,3,4}=> {4,16,64,256} if b and c are 0 then the values a can take will be {1,2,3,4,5,6,7,8} So 2^{1,2,3,4,5,6,7,8} => {2,4,8,16,32,64,128,256} Now 2^a + 4^b + 8^c = 328 Let’s say 2^a = 4^b = 8^c = x => 3x = 328 => x = 109.3333…. Therefore average of the 3 numbers 2^a, 4^b and 8^c is 109.3333…. Clearly this implies that one of the numbers 2^a or 4^b or 8^c has to be greater than 109.333…. 8^c can only take values {8,64} so this means that either 2^a has to take values {128,256} or 4^b has to take the value {256} Let’s say 2^a takes the value 256 then and 8^c is 64; 256 + 64 = 320 now there is no value of 4^b that can satisfy the main equation; 2^a is 256 and 8^c is 8 then 256 + 8 = 264 So 4^b can take the value 64 to satisfy the equation; therefore (a,b,c) = (8,3,1) if 2^a = 128 + i) (8^c = 8) = 136 ii) (8^c = 64) = 192 Let’s take subset in the ranges of b mainly being {64,256} case1: 136 + {64,256} = {200,392} case 2: 192 + {64,256} = {256,448} In both the above cases it is clear that if 4^b was to take the value 256 the equation is not satisfied and if 4^b was to take any value less than or equal to 64 we get the summation of the 3 numbers to BS less than 200 or 256 which still don’t satisfy the equation if 4^b takes the value 256 then 4^b = 256 + i) (8^c = 8) = 264 ii) (8^c = 64) = 320 Case i) 264 + 2^a = 264 + {2,4,8,16,32,64,128,256} = {266,268,272,280,296,328,392,520} Case Ii) 320 + 2^a = 320 + {2,4,8,16,32,64,128,256} = {322, 324, 328, 336, 352, 384, 448, 576} therefore possible solutions of (a,b,c) are: (8,3,1), (6,4,1), (3,4,2)

@shreyesveer8073

Жыл бұрын

Sorry I just realised some typos, in the first line I meant c not x and in the scenario where I explained why 2^a =/= 128 I meant “summation of the 3 numbers to be less than or equal to 200 or 256”

easier method would be realizing that 328 = 256 + 64 + 8 by its binary form.

3 4 2 a b c respectively.

nice and easy

a is 8, b is 3, c is 1.

Sir solve this equation find the equation of circle inscribed in a triangle whose sides are lines line one x+y=8 line two 2x+y=22 line three 3x+y=22

@MathBooster

Жыл бұрын

Suppose the centre of the circle (a,b) Now use the concept : perpendicular distance of centre from all the three lines will be equal, and that distance is radius of the circle.

@ghougamoussah7166

Жыл бұрын

we can solve using the dot product .

I tought about differently i noticed that if you take down 328 to 256 + 64 + 8 * means to the power ok? 2*8 2*6 2*3 and there yo go A= 8,6,3 B= 4,3 C= 2,1 I never solved those type of problesms thats not those kinds X>5 or X+ y =10 and x -y = 20 you know those type? But not simlpe like i did rn just so u get the idea

c=41 after 328/8. a=b=0.

328 = 41 * 8, so we can factor 8 from the left side of the equation and we need to get 41 with powers of 2. Clearly the sum is odd so one of the powers is equal to 1, so either a = 3 or c = 1, then it is just solving the equation for other 2 letters so that the sum is 40, for example: 2^(2b -3) + 2^(3c - 3) = 40. Factor 8 from both sides and then it is easy to see that the RHS = 5, which can only be equal to 4 + 1 aka 2^2 + 2^0. Quick trial and error and the answer follows

My solution (of course tremendously faster): By the definition of the binary writing, there is a unique way to write 328 as a sum of powers of 2. To find it quickly, all we need is to divide 328 by 8, which gives 41, and it's obvious that 41=32+8+1. Therefore we have: 328=2^8+2^6+2^3. Let's discuss on the possible values of c. c can be equal to 1 (fitting with 2^3) or to 2 (fitting with 2^6). If we choose c=1 then b can be equal to 3 (and then a=8) or to 4 (then a=6). If we choose c=2 then b must be equal to 4 and then a=3. Therefore the three solutions are: (8,3,1), (6,4,1) and (3,4,2). Problem destroyed.

@someone-ol8wc

Жыл бұрын

👏👏👏👏so proud👏👏👏👏

@irrelevant_noob

Жыл бұрын

TBF, in the general case it's not unique, since you can still have 16 + 16 + 8 = 32 + 4 + 4...

@italixgaming915

Жыл бұрын

@@irrelevant_noob I didn't write everything but of course you can't have the same power of two more than one time in the sum.

@explanationtomakethisworld9833

Жыл бұрын

@@irrelevant_noob but dear 1 is left which is extra so it doesn't match or must match as powers but having only 3 bases does the work of getting compared and having no way left to miss as their powers must be equal.

@irrelevant_noob

Жыл бұрын

@@explanationtomakethisworld9833 what in the blazes are you rambling about?!? Care you give a specific example to show what your point even is?! o.O

👍👍👍

Technically, aren't the first 2 values you yield out incorrect since their value of 'a' are different than that of 'p', with respect to the answer which had P's value 3

@irrelevant_noob

Жыл бұрын

Not at all, the {a, 2b, 3c} = {p, q, r} is an equality of SETS not of ordered triples. so the positions within each side don't matter.

The answer is zero. ... ... ... You can't arrange 3 numbers to make an ordered pair.

@theeternalsw0rd

Жыл бұрын

If only they had asked for 3-tuples.

You can also substitute the options and verify it. We dont get so much time in the exam.

а=3, в=4, с=2

But how 2^q-p(1+2^r-q)?

Write 328 out in binary gives you a solution in fairly short order.

a=3, b=4, c=2

Another answer is: a = 3, b = 3, c = 8/3 Edit: missed the part where it’s asked to find natural numbers

Sir I try more time I can not solve it please solve dear math booster on youtube I need how to solve

a=3,b=4and c=2

First and solved And can you tell me the GRADE-LEVEL of this olympiad problem

@MathBooster

Жыл бұрын

12

@themangobui1474

Жыл бұрын

Wait what for real?

(3,4,2) is the answer that was got by me.

A is 6 B is 4 C is 1....I did this before seeing the video( solved within 1 minute )..let's see what u've done

🥰🥰🥰

❤️❤️❤️❤️❤️

328=41x8=(1+40)x8=(1+8+32)x8

Attack intuitively. c cannot be greater than 2. Therefore try 2, equalling. 64. 328 - 64 = 264. This is slightly more than 256, suggesting that b may well be 4. If so, then the remainder becomes 8, which makes a = 3. Therefore the answer is a = 3, b = 4, c = 2.

@JohnSmith-nx7zj

Жыл бұрын

That’s not the only solution though.

@notsoancientpelican

Жыл бұрын

@@JohnSmith-nx7zj I understood it was for positive integers--?

@JohnSmith-nx7zj

Жыл бұрын

@@notsoancientpelican You’ve correctly shown that c can not be greater than 2 (since 8^3 = 512>328) and that b cannot be greater than 4 (since 4^5 = 1024>328). But you can’t then just assume that c=2 and b = 4 give the only solution. c = 1 gives another two in positive integers: b = 4, a = 6 b = 3, a = 8

@notsoancientpelican

Жыл бұрын

@@JohnSmith-nx7zj I see, Thank You

A=3.b=4.c=2

a=3,b=6,c=8

solved it in 25 seconds....328 so c can only be 1,2...sum of unit digits has to be 8 ...so c cannot be 1 so c is 2...a cannot be greater than 6 cuz sum of unit digits crosses 8 and b cannot be zero...c = 2...b =4 a =2 is the only possible answer...this seem long but this is themost basic and easily understandable soolution...btw commented this at 0:10...so idk if the same explanation is said in the video

@JohnSmith-nx7zj

Жыл бұрын

c can be 1. c = 1 b = 4 a = 6 c = 1 b = 3 a = 8 are both solutions.

I lose it after 4 steps can you explain more why you use pand q etc

All the women mathematicians . And TRUE

Math in way of philosophy

Indians are very good at math

а = 6 ; b = 4 ; c = 1

1 minute olympiad of 3,4,2.

Wonderful ,sir from which country u r

@MathBooster

Жыл бұрын

India

@adharapuramnavaneeth9925

Жыл бұрын

@@MathBooster me too sir ,from which state

Eu seria um gênio por olhar e saber a resposta quase de primeira? Kkkkkkkk eu simplesmente fui direto pra 2,2,2. Depois 3,3,3. Depois tentei 3,4,2 e deu kk

its 1 minute to solve in excel using solver

how i solved it. 8^3 is too big thus 8^2 is probably it. 4^4 is 256 and +64 keeps it under 328. and look its only 8 away. so 2^3 must be the last value.

Rewrite 328=256+64+8

Obviously not a math olympiad problem.

a=8 b=3 c=1

You are from which country

We can do this by letting. I got it under 2 mins

Do you call (a,b,c) as ordered pair?

@piman9280

11 ай бұрын

As in a pair of trousers for a three-legged man?

3,4,2

a=6, b=4, c=1

@irrelevant_noob

Жыл бұрын

The rub is that this is not the only solution... so you haven't answered the question of how MANY solutions there are. ;-)

2^3+4^4+8^2 = 328

The disturbing lines coming

@MathBooster

Жыл бұрын

That is a feature of KZread, You can turn them off by clicking the cc button in the top right corner of the video.

Maa vuru tata video pettu

@zeroxgammingff9451

Жыл бұрын

Super bro ❤️❤️❤️❤️❤️❤️

@zeroxgammingff9451

Жыл бұрын

Super♥️💓💗♥️💓♥️

بفهم فقط منهج اعدادي المستقله ثانوي لا

Me just casually attempting to do the equation in my head before seeing this I now feel what it's like to be a Disney villain

2.36 I don't understand why it taken 2^p-q. Can anyone tell m

@irrelevant_noob

Жыл бұрын

Watch again, maybe? That's not p-q but q-p... 2:30 and the reason is factoring out a common term of 2^p. Since q >= p, then 2^q = 2^p * 2^(q-p).

@dont1477

Жыл бұрын

@@irrelevant_noob thank you I don't get how it works but If I put numbers it work perfectly

@irrelevant_noob

Жыл бұрын

@@dont1477 aren't you aware of the law of multiplying powers (of the same base)? b^n * b^m = b^(n+m)... Which can be justified by just expanding out what those powers represent...

@dont1477

Жыл бұрын

@@irrelevant_noob thank you for letting me know.You are a god

💬 ❎

No this is not right you taking 2q_r common but 2r-p divided q-p sorry I m not agreeing with you. Wrong

@Raernor

Жыл бұрын

agree, didnt understood that moment too. how it possible?

@zeroxgammingff9451

Жыл бұрын

Yes bro