Germany - Math Olympiad Question | The BEST Trick
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- เผยแพร่เมื่อ 6 ต.ค. 2024
- You should know this approach. Solution
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@higher_mathematics
#maths #math
a=4, b=2
As per question
a+2ab+b=22
Multiply 2 on both the sides3
2a+4ab+2b=44
Add 1 on both the sides3
2a+4ab+2b+1=44+1=45
2a(1+2b)+1(1+2b)=45
(1+2b)(2a+1)=9×5 or 15×3
Let's take 9×5
So,
(1+2b)(2a+1)=9×5
So,
1+2b=9 and2 2a+1=5
2b=9-1 and 2a=5_1
2b=8 and 2a =4
b=4 and a=2( also vice versa)
Find a from source equation
(1) a = (22 - b) / (1 + 2b) (a > 0, b > 0) => b = 1
or
(22 - b) / (1 + 2b) >= 1 (0 < b = 1 + 2b
or
21 >= 3b
or
b
You only need to check b=1,2,3 . At that point b>a
Excellent, as usual. What I learn is that I need to learn to be more creative. 😊
The key is the solutions are integers. Simply solve for a in terms of b.No fancy manipulation necessary. Always look for simple solutions first. In addition,thats applicable to all linear Diophantine equations.
With Diophantine problems, I start by plugging in 1, 2, and 3 for one of the variables. With this problem, a solution was found when I plugged in 2. That this is the only solution is proven by plugging in 3, which results in 3 + 7b = 22. Here, b must be less than 3.
(Keeping in mind that either a or b could be 2)
If you are to try all the possibilities of numbers that give 45 when multiplied together, why not simply set a value for a the solve for b. For example, let a=1, the equation becomes 1 + 2b + b = 22 or 3b = 21 so b = 21/3 or b = 7. Let set a = 2, the equation becomes 2 + 4b + b = 22 or 5b = 20 or b = 20/5 or b = 4. With a = 3, b is not integer. With a = 4, the equation becomes 4 + 8b + b = 22 or 9b = 18 or b = 2. With a = 5 or 6, b is not an integer and with a = 7, b is calculated as 1.
That solution is so hard to think. It is wise that just put randomly some numbers seemed proper
Two answers are possible
If a is 2 then b is 4
If a is 4 then b is 2
Accordingly
2+2.2.4+4
2+16+4 = 22.
4+2.4.2+2
4+16+2=22
I solved in this way: a + 2ab + b = 22 -> b(2a +1) = 22 - a -> b = (22 - a) / (2a +1). b must be integer so I can compute rhis table
a | 2a +1 | 22 -a
--------------------------------
1 3 21 x (22-a divide 2a +1) b=7
2 5 20 x b= 4
3 7 19
4 9 18 x the same solutions switched
5 11 17
6 13 16
7 15 15 x
8 17 14
9 19 13
10 21 12
Good work from you and Mr Giulio. Thanks
a+b = 22-2ab, so a+b is divisible by 2. Therefore, both are even or both odd. a+b positive, so 2ab < 22 and ab < 11. There are very few positive pairs of (a,b) to test that are both odd or both even that and satisfy ab
Continuing your reasoning. Let a≡a' and b≡b' (mod 4), where 0
How did you come up with tricks like multiplying both sides with 2 and then adding 1 to both sides? I mean, it seemed so random what you did and yet effective. Is there any signal of when to use such tricks and which tricks to use?
Same question here.
What I think he did is that he factored a+2ab and got a(1+2b) or a(2b+1) and since there is still one b left over he matched the factor by multiplying both sides by 2 to get the 2b and added 1 to finally match the factor so he can get 2a(2b+1) + 1(2b+1) = 45
And simplified it to (2b+1)(2a+1)=45.
This can also be done to 2ab + b but for the video it was done as a + 2ab.
a(1+2b)+b=22
a=(22-b)/(1+2b)
For each value of b you have a value for a except if 1+2b=0 (if b=-1/2)
So, number of solutions = Infinite.
Isn't it ?
Bro! Only positive integers are asked.
You could do a(1+2b) +b =22. And now you see you want the single b to be 2b+1 so you multiply by 2 and add 1 and then you can factorise.
This is too complicated,it is because a and b are positive integer,just try a equal to 1 to 6, then you can find all answers
At a quick glance if a = 4 and b = 2 then 4+16 + 2 = 22. This gives two solutions: a = 4 and b = 2. a = 2 and b = 4.
What was missing in the beginning of the video was the restrain, two positive integers.
You didn't say at the beginning that you're looking for integer solutions. This simplifies everything!
You can see it at the very beginning. Just watch closely 😊
He did say it at the beginning, both in words and in writing.
He wrote & spoke of positive integers at the beginning. Don't you think this implies integers?
It's big a deal👏🏻👏🏻👏🏻
I have a math olympiad question, can I send it?
Please send it here, we will either help or learn something😊
(2a+1)(2b+1)=45
(a;b)={(1;7),(7;1),(2;4),(4;2)}
There are more answers: a=22 b=0 and simetrical a=0 b=22 The solve idea is very nice and creative.
a and b must be positive
Eccellente soluzione!
Is it called integer Solution?
2a + 1 >= 3 is not true if a (or b) = 0, you missed the 1, 45 pair and 45, 1 pair or (0, 22) and (22, 0) as long as your solution is only non-negative integers
Error. a and b positive!!!!
Surely I'm very thick-headed, but how can 2a(2b+1) + (2b+1) be simplified to (2b+1)(2a+1) ?
Because A * B +A = A*(B +1)...
?
a = 1 and b = 7 ?
And 22, 0. His answers don't work in the original equation .
@@Mofiaczero is not a positive integer
(1,7), (7,1), (0, 22), (22,0) are the only solutions. Your answers don't work with the original equation.
Sorry my friend , but you are wrong, a and b must be positive. He solved it nicely
You have used totally unexplained steps such as multiplying by two and adding one to both sides. These steps seem picked out of the air with no logical reasoning. You do this a lot on your channel and I’ve commented as such in precious videos. It would be very useful for me if you could explain the reasons why you choose each step please.
How to make math difficult
a=2 and b=4...Just be common sense!
Was your fourth step necessary. Wasting our time