for # 1, it reduces to tedious counting and tabulation then, no fun. :) i somehow assumed one needs the constraint of minimum piece or 1 piece each. oh well...
there is a more 'elegant' solution which is not intended for Form 1, e.g. consider the coefficient of x^29 for SUM[(1+x)^a(1+x^2)^b(1_x^4)^c.....]=1/[(1-x)(1-x^2)...(1-x^16)]
9 Kommentare:
我哋曾幾何時都識計架
好似叫GP定AP嘛~
中一點會識, 係唔駛用都計到
點計呀?
one combination, there is only one representation in binary number.
...but one only has to look at the ratio of rightmost 2 triangles, i.e. the ratio of the hypotenus
= (4/5)^2=16/25
sun bin:
for the 1st problem, $1x29 is a combination, $2x14+$1 is another. so the total combination must be greater than 1 ~~
for the 2nd problem, your answer is close. the correct answer should be 16/(16+25)
ohhh yes, i only got the ratio of grey:white.
for # 1, it reduces to tedious counting and tabulation then, no fun. :)
i somehow assumed one needs the constraint of minimum piece or 1 piece each. oh well...
there is a more 'elegant' solution which is not intended for Form 1, e.g.
consider the coefficient of x^29 for SUM[(1+x)^a(1+x^2)^b(1_x^4)^c.....]=1/[(1-x)(1-x^2)...(1-x^16)]
for the 1st problem, there's a simpler solution.
f(2k+1)=f(2k)
f(2k)=f(2k-2)+f(k)
it's a simple recursion.
Kommentar veröffentlichen