F.4 Maths- polynomial+binomial

1. Let f(x) = x^99 + k, as when f(x) divided by (x + 1), the remainder is 1,
so f(-1) = 1, ie
(-1)^99 + k = 1
==> -1 + k = 1
==> k = 2
Using the above result, we have x^99 + 2 = (x + 1) Q(x) + 1,
ie. x^99 = (x + 1) Q(x) - 1
that means, when x^99 is divided by (x + 1), The remainder is -1. So, when 9^99 is divided by 10, the remainder is -1, since remainder is only 0 to 9, so the remainder of this question is (-1 + 10), that is 9.

2. Since (1 + 7)^88 = 1 + (88C1)*7 + (88C2)*7^2 + . . ., except the first term, all the other terms are the multiple of 7. So, when 8^88 is divided by 7, the remainder is only the first term. ie. 1.

3. Since 81^77 = (9^2)^77 = 9^154 = (-9)^154 = (1 - 10)^154
= 1 - (154C1)*10 + (154C2)*10^2 + (154C3)*10^3 + . . .
Same as Q2, except the first 2 terms, all the others are the multiply of 100, so when 81^77 is divided by 100, the remainder is only the first 2 terms. ie. 1 - 154*10 = -1539, as the remainder is only 0 to 99, so the remainder of this question is (-1539 + 100*16), that is 61.

Hope it can help you.

Remainder could not be negative.

1) put x=-1
(-1)^99+ k =1
k=2

(x^99+2) = Q(x)(x+1) +1 (given)
put x=9 9^99 = Q(x)(10) -1
so remainder= -1

2) (1+x)^88 = (1^88 + 88x + 88C2 x^2+... +x^88)
put x=7
8^88 = 1 + 88*7 +88C2 7^2 +...+ 7^88
divided by 7 , remaider =1

3)(1+x)^77 = 1^77 + 77x + 77C2 x^2 +... + x^77
put x = 80
81^77 = 1 + 77*80 + 77C2 80^2 +... + 80^77
= 6161 + 77C2 80^2 +...+80 ^77
when divider by 100 , remaider= 6161= 61(100) +61
so remainder = 61

2012-02-14 21:59:25 補充：
1) k=2
(x^99+2) = Q(x)(x+1) +1 (given)
put x=9 9^99 = Q(9)(10) -1 where Q(9) is constant
9^99 = [Q(9)-1](10) +10 -1
=[Q(9)-1] +9
remainder should be 9

thanks 知識就是力量 !!

2012-02-15 23:14:50 補充：
[IMG]http://i331.photobucket.com/albums/l445/asd256_2008/maths1.png[/IMG]

2012-02-15 23:31:27 補充：
睇唔睇到圖
講多少少
by division algorithm
Let Q(x) be the non-zero polynomial
when Q(x) is divided by (x+1)
可以寫成
Q(x)= q(x)(x+1) + s s係constant,同時係remaider
例如第2條咁
8^88 = 1 + 88*7 +88C2 7^2 +...+ 7^88
= 7(88 + 88C2*7 +...+ 7^77) + 1
所以1係個remaider