A nice and quick elementary number theory problem.
ฝัง
- เผยแพร่เมื่อ 21 ส.ค. 2020
- Using elementary techniques, we solve a quick equation.
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My solution:
a + 6b = 3ab
3ab - a - 6b + 2 = 2
(a-2)(3b-1) = 2
case 1: a-2 = 1, 3b-1 = 2 => a=3, b=1
case 2: a-2 = 2, 3b-1 = 1 => a=4, b = 2/3
since we want a,b € N, the only solution is a=3, b=1
Ha! what I did.
We should consider when a-2=-1 or -2, 3b-1=-2 or -1 respectively, but none of them are answer.
@@Chid417
Agreed, but those cases can be eliminated immediately by noticing that 3b-1 >= 2 (and is therefore positive).
My solution: It's obvious that a must be a multiple of 3 so set a = 3n, and we get
n + 2b = 3nb
after dividing by 3. Some simple algebra give us
2b/n = 3b - 1
LHS ≤ 2b while RHS ≥ 2b, thus the only solution is when we have equality which means b = n = 1 => a = 3, b = 1.
Wow that is clever! (and much simpler)
Very elegant and simple!
"It's obvious" is not a proof, though.
@@sascharambeaud1609 Ok, but we say "it's obvious" in comments and short statements for brevity.
Take (mod 3) of both sides. This shows that a is a multiple of 3
You missed the (0,0) solution when you divided by n
Edit: I realised (0,0) might not be considered a solution depending on if you consider 0 to be a natural number or not
At 7:40, there's quicker way: c = 3b(c-1) means c-1 divides c, but they're coprime, so c-1 = 1, so c = 2. 3 has to divide c, but that's impossible.
I appreciate you recognizing the different definitions for natural numbers after this was pointed out some videos back! You have an obvious dedication to clarity.
9:39
Error in the thumbnail though? a-6b instead of a+6b
With me!
Now u are also providing thumbnail checks. You are really diversifying services 😉
True
thanks! now i know why i've been getting it wrong
A solution to the thumbnail (a-6b=3ab) (trivial solution is a=b=0 and there are no other solutions)
First: we can rearrange this into (-a)(3b-1)=6b so either a is a multiple of 6 or 3b-1 is. It is impossible for 3b to be one more than a multiple of 6 since multiples of 3 are either 0 more or 3 more. So, a is a multiple of 6.
Second: a < 0 since a=3ab+6b=3b(a+2) and if a were positive, a+2 would have greater absolute value, and multiplying by 3b cannot decrease the absolute value to make it equal to a again. Since a is negative, we known b must be positive so we don't change the sign by multiplying.
Third: a >=-3 since (over the reals) a/(3a+6)=b. Note that (3a+6) != 0 since a must be a multiple of 6 and -2 is not a multiple of 6. b >= 1 because it is a strictly positive integer. So a >= (3a + 6) if 3a+6>0 (which is impossible since a is a negative multiple of 6) or a = -3. There are no mulitples of 6 between 0 and -3 so the trivial solution is the only solution
Nice puzzle. Another way to do it is to rewrite a + 6b = 3ab as (a - 2)(3b - 1) = 2. So a - 2 is either 1, -1, 2 or -2. Checking those gives you all the solutions: (a,b) = (0,0) and (3,1).
This was my method also.
Nice and simple, I like this.
Don't need to mess about with gcd if you rearrange it to (a-2)(3b-1)=2. 3b-1>=2, a-2>=-1, so the only solution is 3b-1=2 and a-2=1.
My solution:
a = 3ab - 6b
a = 3b * (a-2)
a/(a-2) = 3b
since b >= 1:
a/(a-2) >= 3
a >= 3a - 6
0 >= 2a - 6
a
Actually, u know that a-2 can't be 0 and must be a positive number since we are on natural numbers domain. So, a must be bigger than 2 and the only case u have to check is a = 3 and it gives the right solution.
@@clarksoncastro2218 That is incorrect. a-2 can be 0 or a negative integer in this problem.
Yes, makes sense,but you are assuming a>2 when multiplying by a-2 in a/(a-2)>=3 - - > a>=3a-6. You can deal with the case a-2
@@wospy1091 but we are dealing with natural numbers
@@timetraveller2818 Just because you have a natural number domain doesn't mean you have a natural number range. AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAHHHHHHHHHHHHHHHHHHHHHHHHH
Surely, pointing out that the gcd of n and n+2 cannot be greater than 2 is a matter of pointing out that multiples of 3 are 3 apart, etc.
Hey, no fair! In the thumbnail it said a - 6b = 3ab, and you're solving a + 6b = 3ab.
In solving the thumbnail, I did pretty much what Adam Romanov did, but with the different sign.
3ab + 6b - a = 0
(3b - 1)(a + 2) + 2 = 0
(1 - 3b)(a +2) = 2
I then drew a graph of a against b to find all solutions (within the scope of the graph)
Could you start a couple of series on 1) the Calculus of Variations & 2) Integral Equations?
From a = 3b(a-2), we have that a-2 divides a, but that means a-2 is at most half of a i.e. a is at most 4.
Even better, since we know the gcd of a and a-2 is 1 and a-2 divides a, a-2 must be one because if the smaller number divides the bigger one, it is their gcd.
Hence a=3. Good catch man!
If the above question given as a subjective one . We have to solve question as suggested by michael sir. But if it is objective one we can solve like this :
Let a = x & b = y
Then we have x + 6y =3xy
So, 3y = (x)/(x-2) ...........(i)
As x,y € N y>0
So (x)/(x-2)>0
so x € (-infinty , 0 ] U (2 , + infinty ) as x € N so x € N - {1,2}
From (i) , 3y = (1+ (2)/(x-2))
As x € N - {1,2} x-2 € N - {1,2}
So x-2 l 2 x = 3 y = 1 is the solution for the above eq.
So a=3 ,b = 1 are only sol.
But michael sir sol is ultimate for subjective questions.
Much easier way: (a-2)*(3*b-1)=2, then abs(3*b-1)
a=b=0 also works. That's why you have to split into multiple cases like he usually does in these videos.
@@jadegrace1312 zero is not a natural number on this channel, that is why he hasn't found that solution. Spending 10 minutes for a totally trivial problem, not solving in the standard and much faster way is too much waste of my time.
@@robertgerbicz the lemma is more important than the question
We can solve it using modular arithmetic as well. Integral solutions are (0,0) and (3,1)
One could also "complete the product" as you have dubbed it before, and garner the same result using the fundamental theorem of arithmetic (unique prime factorization).
And one more solution:
Solve for b :
b=a/(3a-6)
and check for natural solutions.
a=0 → b=0
a=1 → b
I find it interesting that all five of those values of “a” give a “b” value in a different family: A whole number, a negative integer, a non-number (or should I call it THE non-number?), a natural number, and non-integer rational numbers.
When we express the form like a/(3a-6), we should separate the cases whether the denominator is 0. In this problem, that is when 3a-6=0.
However, in the result, the answer is same in this problem. The denominator is zero means that some number times 0 is 2, so it carries out same answer.
Here is my procedure:
a+6b=3ab
Taking everything mod 3, we have
a+0=0
Clearly, a=0 mod 3, so a=3c, where c is a natural number.
We have that
3c+6b=9cb
c+2b=3cb
2b=3cb-c
2b=c(3b-1)
c=2b/(3b-1)
We have that 2b/(3b-1) is a natural number. If this quantity is less than 1, it clearly can't be. To find an upper bound for b, we set
2b/(3b-1)
Tks bro
When you said a must divide RHS so a|3b you can also say 3b must divide LHS so 3b|a thus a=3b and replacing this in the equation to solve a then solve b
.
3 divides a clearly. a=3k. k+2b=3kb and k=(3k-2)b. Hence 3k-2 less then or equal to k. Hence k=1 and a=3. So b=1 which gives (a,b)=(3,1)
Another solution: we know a and b are integers and a > 0 and b > 0. We start with
a + 6 * b = 3 * a * b
Divide by a * b to obtain
1 / b + 6 / a = 3
1 / b is bounded as
0 < 1 / b
rearranging you get a=3ab-6b, therefore a must be divisible by 3 so substituting 3k for a you get 3k= 9kb-6b and b=k/(3k-2) if b is a natural number |k|
Hi. I think it would be good to clarify that gcd means greatest common divisor. In the
UK we call it highest common factor HCF so it threw me for a bit what gcd stood for. I had to google it.
Then in UK, do you use term LCF (Lowest Common Factor) instead of LCM (Least Common Multiplier)?
Thanks! I was thinking about why he'd bring up the great circle distance for this one 😜
I am touched
The sign was flipped in the thumbnail!, oh no
Here is my solution today.
a + 6b = 3ab, obvsly a is some 3n, so let a = 3n and factor out 3
n + 2b = 3nb, isolate n and n = b(3n-2), i saw the hint and got the idea immediately, multiply both sides by 3 to get 3n = 3b(3n-2), isolate b and 3b = 3n/(3n-2), which is also 1 + 2/(3n-2), this means n = 1 or n = 4/3, so a = 3 or a = 4. But since the left side says 3b, n cannot be 4/3, so n = 1, and b = 1, so we have
A = 3 and B = 1
Yeah, he got me too, had to solve it twice!
I solved like this: a+6b=3ab => a|(a+6b) => a|6b => 6b=ak for some integer k. Similarly, b|a => a=bl for some integer l. Putting this in prev eqn we get kl = 6 k and l are natural number. So we can find all ordered pairs and then solve.
a = 3b(a - 2) implies a-2 | a and 3 | a. From Euclid's algorithm we know that any common divisor of n and n+m divides m, so is ≤ m (if m > 0), so (taking n = a-2, m = 2), a-2 ≤ 2, so 1 ≤ a ≤ 4, so a = 3, b = 1.
Even shorter, (a-2) is a factor of a since a = 3b(a-2) but gcd(a, a-2) = 1 implies that a-2 = 1 and thus a=3
@REEEeE Please How do you determine that gcd(a, a-2)=1? Couldn't it also be 2 (if a is even) or 0 (if a is 2)? Also, if the gcd is 1, don't we still need to check the case where a-2 is -1, not 1, because in that case we also have gcd(a, a-2)=1?
@@zygoloid this occured in the case that a is odd and thus gcd = 1
And gcd cant be 0 since 1 still divides 0
Negatives cant occur since we deal with natural numbers (b would be negative)
We can write this as a+6b=1.5ab+1.5ab. 1.5ab is always greater than a over the natural numbers, so if 1.5ab is greater than 6b this equation has no solutions. 1.5ab>6b is the same as a>4. Therefore we only need to check the values a=1,2,3,4. Easy casework at this point, only a=3 gives us a solution over N
b = a / (3a - 6). If a > 3, a < 3a - 6 so b
a = 3ab - 6b
b = a/(3a - 6)
for b E N clearly a >= 3a - 6; 6 >= 2a hence a < = 3
Trial and error with a = 1, 2 and 3 yields a = 3 (and b = 1) as the only answer
Another solution:
It is obvius that a = 0 mod 3 so set a = 3n and we get:
n + 2b = 3nb
Again it is obvious that n = 0 mod b so set n = bm and we get:
m + 2 = 3bm ---> m(3b-1)=2
Because 3b - 1 is an integer m must divide 2 also m is a natural number so we have either m = 1 or 2. Trying these two values, we get a = 3 and b = 1 as the only solution.
Move everything to one side, so we get 3ab - a - 6b = 0. Now add 2 to both sides: 3ab - a - 6b + 2 = 2. The left side factors, giving us (3b-1)(a-2) = 2. Since the left side is the product of two integers, it must be the case that this product is 1*2, 2*1, (-1)*(-2), or (-2)*(-1). Only the second and third of these yield integer values for b, giving us either b = 1 (and a = 3) or b = 0 (and a = 0). So the solutions (a,b) in integers are (0,0) and (3,1). Depending on whether N includes 0 for you or not, the first might not count as a solution for you.
Edit: Oh, I see someone else also commented this solution. Nice!
When doing the case analysis, when a is odd, as soon as you have a=3b(a-2), immediately gcd(a,a-2)=a-2, so using the lemma, one has 1=gcd(a,a-2)=a-2, so a=3. In the even case, once you have c=3b(c-1), it's clear that 1=gcd(c,c-1)=c-1, so c=2. I think this is faster and more clear than introducing new factors m and n, and making substitutions.
Well when we have:
a = 3b * (a-2), we can just say that a is a multiple of a-2.
But no a bigger than 4 can be divided by a-2 because 1
My solution:
move everything to the left side and factor to get (2-a)(3b-1) = -2 and now you have that -2 can be expressed as (-2)(1), (2)(-1), (-1)(2) and (2)(-1) because a and b are whole numbers so are (2-a) an (3b-1) also. checking the four cases you only get (3,1) as the solution.
My solution:
Substitute A and B for X and Y
x + 6y = 3xy
Subtract both sides by the 6y term
x = 3xy - 6y
Factor out the common 'y'
x = y(3x-6)
Divide both sides by the 'y' term
x/y = 3x - 6
Take the reciprocal of both sides
y/x = 1/(3x-6)
y/x can be interpreted as the slope of a function. When x = a value, y = a value (and vice versa)
Therefore, we know that y = 1
x can be solved for via the denominator of each side
x = 3x - 6
Subtract over the 3x term
-2x = -6
Divide both sides by -2
x = 3
Answer: x = 3; y = 1
How would u do it in C ?
His voice is so soothing
His handwriting is better than most of my teachers
are you stating that his handwriting is better than your teacher's handwriting or that it's better than your teachers themselves?
@@lakshaygupta9061 handwrting is better obv
@@lakshaygupta9061 sorry made a grammarical error
Premium quality Hagoromo chalk helps
a+6b=3ab
=> a=3b(a-2)
So either a=b=0 or a is divisible by 3, b and (a-2)
a is divisible by a-2 => a-2 has to be 1 or 2 => a=3 or 4
since a is divisible by 3 => a=3, hence 3=3*b*1 => b=1
So the solutions are (0,0) and (3,1)
"a is divisible by a-2 => a-2 has to be 1 or 2"
Careful, a-2 can also be -1 or -2.
@@zygoloid You are right, thanks! Luckily this does not change the flow afterwards.
3b = a/(a-2). So only a = 4 and a = 3 will yield an integer at the RHS.
But only when a = 3, we get an integer for b, so b = 1.
And that's a good place to stop.
Alternative :
3|a and so let a=3a1 so the equation transfer to a1+2b=3a1b and then for a,b>=2 LHS
5:00 Doesn't it have to be 3b|a, since a is the product of the two smaller terms 3b and (a-2)?
Same doubt
Yes, 3b|a and (a-2)|a by the very definition of multiplication, but that's not what he's talking about here. He should have elaborated this a little bit, but here it goes: by identity property, a number divides by ITSELF, so a | 3b(a-2) and 3b(a-2) | a.
Focus on the first one.
a divides by the product of the factors, but since he established that a and a-2 are coprimes, which means they can't divide without a remainder, you must conclude that a | 3b. That only happens because we know that a and a-2 are coprimes.
Yes you're correct. And since he proved a|3b, we know both the numbers divide each other. And since both are positive, they can't be the negative of each other and so have to be equal (please prove this, I can't)
So we can cancel the a on the LHS and the 3b on the RHS and get a-2=1 which gives us a=3 and therefore b=1
An alternative way basically.
Edit: Oh wait I got it.
If a|m and m|a, then we know ak=m and mn=a. Substituting one into the other, we get kn=1 and since they're both >0, they're both 1 and so a=m
@Jamirimaj Then, doesn't this immediately imply that a-2 is equal to 1?
1:25 gcd(n,n-2)=gcd(n,n-(n-2))=gcd(n,2)=1or2 using gcd(m,n)=gcd(m,m-n)
a = 3b * (a - 2) and b>=1 => 3*(a-2) 2*a - 6 a
Hmm. SInce (a-2) is a factor of a, then gcd(a,a-2)=a-2. Just set a-2=1 or a-2=2 and get a=3 or a=4. Then since 3|a, then a cannot be 4. So a=3.
I think this answer isn't quite complete. gcd(a,a-2)=±(a-2). So you also need to rule out a-2=-2 and especially a-2=-1. For example, your argument would also apply to a = (3b-4)(a-2) but that equation has a solution with a=1.
@@zygoloid Thanks for pointing this out. I did not consider these cases since a is a natural number.
But as you pointed out, a-2=-1 will give a=1 which is still a natural number. So this case is still worth checking. Thanks again!
i would like to try this but for (a-3) instead of (a-2).
so we look at gcd(a, a-3):
a and a-3 may only share the factor 3, if anything, for the reason c and c-1 are always coprime.
then we look at the cases:
presume a = 3c for some natural c.
we have 3c = 3b(3c-3), then divide by 3 to get c = 3b(c-1). c and c-1 are coprime, and so 3b = mc, so 1 = m(c-1), c=2, m=1, 3b=2 which is a contradiction. a is never a multiple of 3.
presume a = 3c+k, where k is 1 or -1
we have 3c+k = 3b(3c-2) -> 3c+k = 9bc-6b -> k = 3(3bc-2b-c), which is a contradiction since k is not a multiple of 3.
in conclusion this necessitates that a+9b=3ab has no non-zero integer solutions.
next we can look at a+6b = 2ab. here the factoring is a=2b(a-3), where again we have the cases...
a = 3c -> c = 2b(c-1), c = 2, and b = 1.
a = 3c + k -> k = 6bc - 6b - 3c, which is a contradiction.
the only solutions here are (6,1) and (0,0)
bruh it was much simpler... either SFFT like in adam romanov's solution, or you go like this:
if one among a and b is 0, the other is too, and we get (0,0) as a solution.
if a,b>0 you get that b divides a which divides 6b, so a is either b, 2b, 3b, or 6b.
the four cases give 7=3b, 8=6b, 9=9b or 12=18b. the third option is the only possible one, which gives (a,b)=(3,1)
There is a simpler way to solve. write the equation as 1/b +6/a =3 (assuming a,b non-zero). Now, 1/b can maximum be 1 and 6/a maximum is 6 for a=1. For a=2 we have 6/a=3 and for a =3 it is 2. We can not have 'a' bigger than 3 because 1/b can never compensate to make the sum =3. So we just have to test with these values. But 'a' can not be 1 or 2 because 6/a is bigger than equal to 3 (so adding positive 1/b will never make it equal to 3). So the only case you really need to test is a=3. Does there exist a b, when a=3 such that 1/b +6/a =3. yes, b=1.
Yes but you're missing a=b=0
@@jadegrace1312 No. As said in the video, you immediately see a=b=0 is a solution. Then you assume if a and b are non-zero, you can divide by ab. Also, 0 might be excluded from the set of natural numbers in some conventions.
@@lavneetjanagal that makes sense
gcd(n,n-p)=p if p|n, 1 else, for n element Z and p prime.
6b=3ab-a
6b=b(3a-a/b)
6=3a-a/b
6=a(3-1/b)
(3-1/b) must be natural so only solution is b=1, and then a=3
it wasn't my first method of thinking, i tried to think in limits of a or b going to infinite which gives the other parameter under 1. then i arrived accidentally to creating a key element of '1/b'
3ab-6b-a=0. Iff 3b(a-2)-a=0. Iff 3b(a-2)-a+2=2. Iff 3b(a-2)-(a-2)=2. Iff (3b-1)*(a-2)=2. It follows 3b-1= +-2 or 3b-1=+-1. But 3b-1 is congruent to 2 and -1 mod 3. Therefore, if 3*b-1=+-2 then 3*b-1=2, b=1 and a-2=1. i.e. a=3 and b=1. Now, if 3*b-1=+-1. Then 3*b-1=-1. Therefore, b=0. a-2=-2, i.e. a=b=0.
Odd case: The equation shows a-2 is a factor of a, but as gcd=1, a-2=1, then a=3. Similarly, in the even case a-1 is a factor of a, so a-1=1, a=2. In the even case, of course, b is not an integer.
That means that
3ab - 6b - a = 0
3b(a - 2) - a = 0
3b(a - 2) - a + 2 = 2
3b(a - 2) - (a - 2) = 2
(3b - 1)(a - 2) = 2
That means that 3b - 1 can only be -2, -1, 1 or 2. However, since b >= 1 (because it is natural), then 3b - 1 >= 2, so 2 is our only candidate for 3b - 1, and therefore a - 2 = 1
3b - 1 = 2 implies b = 1, and a - 1 = 2 implies a = 3.
Our only solution is a = 3, b = 1.
I found one way.
a+6b=3ab
Taking mod 3 on both sides
a=0 (mod 3) [I couldn't find the "three parallel lines" thing]
a =3n
One obvious solution is (a,b)=(3,1)
Let a>3
6b=3ab-a= a(3b-1)>3(3b-1)=9b-3
So 6b>9b-3
This is not true for any b>1(b is a natural number)
So a cannot be greater than 3. But a is a multiple of 3. So no other solutions other than (3,1) exist
These mod solutions never fail to blow my mind. Clever way!
It's a bit complex decision, I think. ax+bxy+cy=d equal (bx+c)(by+a)=bd+ac. That means a+6b=3ab equal (3a-6)(3b-1)=6 and so on...
If you know that a=f(b) is a a=k/b function that was "mooved" then you can draw in seconds this function, by knowing its asimptotus, and one point is enough... then the answer is pretty obvious.
My solution
a+6b=3ab
=>a=3b(a-2)
=>3b=a/(a-2)
Since b is a natural number, minimum value of 3b=3
Note that the RHS is a strictly decreasing function & that a=4 gives a/(a-2)=2 which is less than 3
So a is one of 1,2,3
If a is 1, a/(a-2) becomes negative which makes b negative which is not possible since it's a natural number.
a=2 makes the denominator zero so doesn't work either
Hence a=3 gives the only solution. Thus 3b=3/(3-2)=3 => b=1
Thus, *(3,1)* is only solution
Can i just factor this like (3b - 1)(a - 2) = 2 and then factor 2 like 2*1 since we are in natural numbers domain? This way we see that the only solution is a = 3 and b = 1.
You can factor that way. Michael tends to use the nuclear approach to swat a fly, but that's what makes his videos interesting 😁.
Another solution:
a+6b=3ab => a = 0 mod 3, so a = 3k.
3k+6b=9kb => k+2b=3kb => 2b = 3b mod k => 2 = 3 mod k since b is never 0. This only holds for k = 1.
So k = 1 => a = 3, b = 1.
Equation can be rewritten as 6b/(3b-1)=a because 3b-1 =/ 0 for all integers.
Assume a > 2.
6b/(3b-1)>2
=> 6b>6b-2
=> 0>-2
=> there are no solutions for a > 2
Check a =1 and we get the solution (1,3).
Wow this is just awesome
awesome
I used factorization after subtracting 2 on each side
Here is a nice and quicker solution: 3b = 1+2/(a-2), then obviously a=3 b=1 is the only solution.
Given a and b are natural numbers.
Given a + 6b = 3ab.
So, a = 3ab - 6b = 3b(a - 2).
So, (a - 2) | a, but this is only possible if a = 3.
Now a = 3 & a + 6b = 3ab, then 3 + 6b = 9b.
Hence, a = 3 and b = 1 is the only solution
to a + 6b = 3ab
5:14
Error I think : case 1: a= 3b(a-2) then 3b|a => a = 3mb case 2 : 2c = 3b(2c-2) => c=3b(c-1) => 3b|c => c=3nb
a/3b with Gauss theorem.
a=3b(a-2)
if a=2, a=0. This is false, so a!=2.
a/(a-2)=3b
If a=1, b2, so a>a-2>0.
LHS>1, so LHS>=2.
Since LHS decreases as 'a' increases, 3
Another solution :
a = 2 gives nothing interesting (all b is valid), so we can exclude a = 2.
Notice that with a =/= 2, from a + 6b = 3ab we get b = a/(3a-6).
The function x -> x/(3x-6) :
1. Takes value 0 in 0, so we get the first obvious solution.
2. Is negative over the positive integers strictly less than 2, so no positive solution there.
3. Has no value for x = 2, so no solution there.
4. Is strictly between 0 and 1 for all positive integers strictly superior to 3, so no positive integer solutions there either.
5. Hence, the only strictly positive integer solution possible is for x = 3.
Then look at a = 3, you get b = 1, and you have proven that it's the only strictly positive integer solution.
And that's a good place to stop.
Very simple: 6b=a(3b-1). Since 3b and 3b-1 has no common divisors, 3b-1 must divide 2, so 3b=3 (3b=2 is not possıble) that is b=1 and a=3.
How can we say for sure that p-q isn't equal to 2 instead of d? I think some details are missing or I just don't get it
We can't - that's why d can also be equal to 1. But we can say that 2 is a multiple of d, which is what was done here.
He didn't say d=2. He said d|2 which allows d to be both 1 and 2
I solved for b: b = a / [3(a - 2)].
For b ∈ N, a ≥ 3(a - 2) → a ≥ 3a - 6 → a ≤ 3.
Because a ∈ N, if a = 1, b = -⅓; If a = 2, b is undefined.
If a = 3, b = 1. This is the only solution.
Is the error in the video thumbnail an intention one ?
The known condition is also (a-2)(3b-1)=2, a,b∈N. So easy to get (a,b)=(3,1). But not as natrual as Prof Penn's method.
Great
@4:50 No. 3b | a
This solutions was unnecessarily overcomplicated. Considering that for large a,b the product ab is much larger than a+6b, one can simply argue that for a>=4 and b>=1 the RHS is 3ab=ab+2ab>a+6b, so it is not a solution. Also considering that two of the three terms are multiples of 3 gives that in a solution "a" must be a multiple of 3. So a=0, a=3, or b=0, leading to the solutions a=b=0 and a=3, b=1.
Your solving method is awesome. Why don't you write books?
a + 6b = 3ab... so 0 = 3ab - a - 6b, then 2 = 3ab - a - 6b +2.
So... 2 = (a - 2)(3b - 1).
Because a-2 and 3b -1 are integers... then we have 4 cases:
a-2 = -2 and 3b -1 = -1, here a = b = 0
a-2 = -1 and 3b -1 = -2, here a = 1 and b = -1/3 (we discard this solution because B is Not an integer)
a-2 = 1 and 3b -1 = 2, here a = 3 and b = 1
a-2 = 2 and 3b -1 = 1, here a = 4 and b = 2/3 (we discard this solution because B is Not an integer)
Another solution could be
a = 3ab - 6b
a = 3b (a-2)
a/(a-2) = 3b
(a-2+2)/(a-2) = 3b
1 + 2/(a-2) = 3b
2/(a-2) = 3b-1.
RHS is a natural Number, so 2/(a-2) must be a natural Number.
Here we have that a-2 must be 1 or 2 and check cases
Long your solution: here mine
a/b+6=3a so b divide a; a=k*b give k+6=3kb; 6=k(3b-1) the only solution is k=3 and b=1
Completing the product(forcing a factoring by grouping) is effective:
a + 6b -3ab=0 , a - 3b(a - 2)=0, (a-2) - 3b(a-2) = -2
(a - b)(1 - 3b) = -2 and equating factor pairs gives a=3, b=1
Seen this idea called SFFT or Simon's Favorite Factoring Trick
Why not factorise (a-2)(3b-1) = 2?
Then the solution drops out straight away.
So much easy
a+6b=3ab
(a-2)(3b-1)=2
b>0 so b=1 and a = 3
solved before watching the solution :D
I said zero is not a natural number because a natural number must occur in nature . We can see 3 trees, 6 apples and say 2 ideas. But we can't see or experience zero trees etc. We can only derive it as an extraction.
I saw the thumbnail and solved that case in Z.
(a,b)=(0,0) or (-3,1)
Good job!
actually i did with hit and trial
Where did you set the limits for the set of natural numbers you were trying, and why did you set it where you did? Or did you mean you tried some numbes and fond an *a* and *b* that worked?
another solution:
a+6b=a(mod b), 3ab=0(mod b) => a=pb, p is a natural number, (p+6)b=3p(b^2) => b=0(and a=0) or p+6=3pb, b=(p+6)/3p, b is >=1(we already find solution with b=0), so (p+6)/3p>=1 => (p+6-3p)/3p>=0 => (3-p)/p>=0 => 0 4=3b, no solution
p=3)9b=9b^2 => b=1, a=3
answer: a=0,b=0; a=3,b=1
Lol thumbnail says a - 6b = 3ab
Me: Hmm so that’s impossible
Actual problem: a+6b = 3ab
Ohhh I just got baited
Not impossible, a=-3 b=1 (as well as the trivial solution)
jesboat natural numbers??
@@zerosumgame9071 oops
Easy problem requires complicated solution...
The icon that introduces this video has a minus sign in the equation, not a plus sign. This would make the problem insolvable.
It is solvable you just don't get any solutions apart from the trivial 0,0
Zero is not a natural number. That's just how I was raised.
Was it just me, or was the audio a little too quiet? Regardless, great video! :)
just sfft
You don't count 0 as a natural number? If you did, a = b = 0 would be an obvious solution.
First!
Zero is not a natural number
In many countries around the world, it's natural to have nothing, hence 0 is part of |N