2^x (4 - x) = 2x + 4 Dividing both sides by 2 gives: 2^(x - 1)×(4 - x) = x + 2 Note that x cannot be odd except 1 (and 1 is a solution), now we put x = 2k and cancel by 2: 2^(2k - 1)×(2 - k) = k + 1, so: (k + 1)/(2 - k) > 0, which gives the boundaries: -1 < k
Question: is there any way for us to know what the question's quickest way to do for number theory questions? sometimes checking positivity is fine, but sometimes it requires mod to do so. And about how long do you need to take for you to think of an elegant solution? thanks
After solving ample problems you can find that you can mix different methods and come with your own. Like in chess some moves can be rejected just by intuition.
The function f(x)=(2x+4)/( -x +4) is a mooved hyperbulla y=k/x so it is very simple to draw it [y=-2 horizontal ... x=4 vertical ...] Now we have only x=-1 , x=0 , x=1 , x=2 , x=3 as possibilities.
You can also do the following. (2^x)(4 - x) = 2x + 4 Since x is integer => 4 - x | 2x + 4 => 4 - x | -2x - 4 => 4 - x | 2(4 - x) - 12 => 4 - x | 12 Or 4 - x = 1, 2, 3, 4, 6, 12. => x = 3, 2, 1, 0, -10, -16. Recheck: x = 3 => 8(1) = 10 (false) x = 2 => 4(2) = 8 (true) x = 1 => 2(3) = 6 (true) x = 0 => 1(4) = 4 (true) x = -10, -16 => LHS is not integer. In conclusion: x = 2, 1, 0.
@@nicklarry7791 Ah, right. This. A possible solution may be to set x = -y, then y > 0 RHS changes very little, but LHS has a big change: 2^(-y)(4+y) Rewritten it as (4 + y)/(2^y), and since RHS is an integer, LHS must be an integer as well, so we're back to finding y satisfying 2^y | (4 + y) for positive integers y
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2^x (4 - x) = 2x + 4
Dividing both sides by 2 gives:
2^(x - 1)×(4 - x) = x + 2
Note that x cannot be odd except 1 (and 1 is a solution), now we put x = 2k and cancel by 2:
2^(2k - 1)×(2 - k) = k + 1, so: (k + 1)/(2 - k) > 0, which gives the boundaries: -1 < k
Your answer is way more straightforward than the video
Question: is there any way for us to know what the question's quickest way to do for number theory questions? sometimes checking positivity is fine, but sometimes it requires mod to do so. And about how long do you need to take for you to think of an elegant solution? thanks
After solving ample problems you can find that you can mix different methods and come with your own. Like in chess some moves can be rejected just by intuition.
No.
There is only one way: develop your profilaxis in maths
0 and 1 are always good first solutions to check.
I divided both sies by 4 - x and simplified the RHS to -2 - 12/(x - 4), which has to be an positive integer. Its easy to narrow down cases from there.
This one is a beauty. Simple but interesting.
X can't be negative because the left part will be fractionary and the left part integer. So X is always positive, no need to test the value -1
of the 5 cases you can eliminate more by simply looking at original equation mod 2
you have problems when x < 0, because 2^x is not invertible modulo 2
The function f(x)=(2x+4)/( -x +4) is a mooved hyperbulla y=k/x so it is very simple to draw it [y=-2 horizontal ... x=4 vertical ...] Now we have only x=-1 , x=0 , x=1 , x=2 , x=3 as possibilities.
Is there any solution if x is not an integer?
You can also do the following.
(2^x)(4 - x) = 2x + 4
Since x is integer => 4 - x | 2x + 4
=> 4 - x | -2x - 4
=> 4 - x | 2(4 - x) - 12
=> 4 - x | 12
Or 4 - x = 1, 2, 3, 4, 6, 12.
=> x = 3, 2, 1, 0, -10, -16.
Recheck:
x = 3 => 8(1) = 10 (false)
x = 2 => 4(2) = 8 (true)
x = 1 => 2(3) = 6 (true)
x = 0 => 1(4) = 4 (true)
x = -10, -16 => LHS is not integer.
In conclusion: x = 2, 1, 0.
2^x = (2x-8+12)/(4-x)
= -2 + 12/(4-x)
indubitably, 4-x>0,
by that, you could limit the choices into only 3
This only works for non-negative integers. Otherwise, 2^x is not an integer, which you're assuming.
@@nicklarry7791 x is *never* smaller than 0, otherwise the LHS becomes a fraction, while the RHS is an integer
@@gdtargetvn2418 That also needs to be proven. What if the fraction resolves to an integer?
@@nicklarry7791 Ah, right. This.
A possible solution may be to set x = -y, then y > 0
RHS changes very little, but LHS has a big change: 2^(-y)(4+y)
Rewritten it as (4 + y)/(2^y), and since RHS is an integer, LHS must be an integer as well, so we're back to finding y satisfying 2^y | (4 + y) for positive integers y
Solide!
Seconds. Also this video: 5 minutes
Yes ... too easy :))
easiest question this month :)
nice
U can also solve it using graphs
2 to the power of x and 2x+4/4-x
Now the points where they intersect those are the solutions
yay i did it
can someone help me with this one:
find the largest value of natural number n such that 8^100!-1 is divisible by 7^n
Factor 8^(100!)-1. You get one factor of 7 and the you just need to look what happens in the other term.
@@jölwritesmusic do you have any ways to factor that large number bc i cant think of any
This problem is not worth a video
So poor!
brilliant way