A polynomial f(x)=y with integer coefficients passes through points (3, 9) and (8, k). Find the largest possible value of k if k<9.
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A polynomial f(x)=y with integer coefficients passes through points (3, 9) and (8, k). Find the largest possible value of k if k<9.
humm, i don't think there is a largest possible value :p
if k<9 there is probably a largest value, and its probably 4
yeah, i think i misread it :S
i am just speculating about (8,4) what do you think kedaman
(k-9)/(8-3) = (k-9)/5
To be integer k is 4. Hehe , assume its linear, :( I miscalculated on the contest.
Don't listen to Sprite and assume it's linear; there's actually a way to do the question and get 4.
there's probably something terribly wrong with my proof trough induction, that theres no integer constants a0 to an that gives a k between 9 and 4, but i could be wrong as well :p
p(n):
9 = a0 + a1*3^1 + .. + an*3^n
4 < a0 + a1*8^1 + .. + an*8^n
9 > a0 + a1*8^1 + .. + an*8^n
prove p(1) is false:
9 = a0 + a1*3
4 < a0 + a1*8
9 > a0 + a1*8
4 < 9-a1*3 + a1*8
9 > 9-a1*3 + a1*8
4-9 < a1*5
9-9 > a1*5
-1 < a1
0 > a1
there's no integer a1 between 0 and -1.
prove p(n)=>p(n+1) is false:
9 = a0*3^0 + a1*3^1 + .. + an*3^n + a(n+1)*3^(n+1)
4 < a0*8^0 + a1*8^1 + .. + an*8^n + a(n+1)*8^(n+1)
9 > a0*8^0 + a1*8^1 + .. + an*8^n + a(n+1)*8^(n+1)
9- a(n+1)*3^(n+1) = a0*3^0 + a1*3^1 + .. + an*3^n
4- a(n+1)*8^(n+1) < a0*8^0 + a1*8^1 + .. + an*8^n
9- a(n+1)*8^(n+1) > a0*8^0 + a1*8^1 + .. + an*8^n
- a(n+1)*3^(n+1) = - 9 + a0*3^0 + a1*3^1 + .. + an*3^n
- a(n+1) = (- 9 + a0*3^0 + a1*3^1 + .. + an*3^n )/3^(n+1)
4 + (- 9 + a0*3^0 + a1*3^1 + .. + an*3^n )(8/3)^(n+1) < a0*8^0 + a1*8^1 + .. + an*8^n
9 + (- 9 + a0*3^0 + a1*3^1 + .. + an*3^n )(8/3)^(n+1) > a0*8^0 + a1*8^1 + .. + an*8^n
and the assumption p(n) could be written as
4 < 4 + (- 9 + a0*3^0 + a1*3^1 + .. + an*3^n )(8/3)^(n+1) < a0*8^0 + a1*8^1 + .. + an*8^n
9 > 9 + (- 9 + a0*3^0 + a1*3^1 + .. + an*3^n )(8/3)^(n+1) > a0*8^0 + a1*8^1 + .. + an*8^n
0<(- 9 + a0*3^0 + a1*3^1 + .. + an*3^n )(8/3)^(n+1)<0, obviously not true
ok let a general function with integer coeeficients:
f(x)=a0x^n+a1x^(n-1)+...+an
f(3)=a0*3^n+a1*3^(n-1)+...+an
f(8)=a0*8^n+a1*3^(n-1)+...+an
notice that 8^k-3^k is divisible by 5 (hint mod 5) so f(8)-f(3) is divisible by 5. go from there
Induction...interesting...but why would you try to prove that it can't be less than 4 or greater than 9, unless you were fairly sure that it would work already, which, on a 30-minute math contest you wouldn't be?
it wouldn't have to be 4, its just that 4 and 9 gave these
4-9 < a1*5
9-9 > a1*5
:)
aah, took me a while to get what you were on, if the terms in the polynom are divisible with 5 then it wouldn't be possible to have k less than 5 from 9Quote:
Originally posted by bugzpodder
notice that 8^k-3^k is divisible by 5 (hint mod 5) so f(8)-f(3) is divisible by 5. go from there [/B]