claywatts

1) Find the remainder when 7^7000 + 7^700 + 7^70 +7 is divisible by 8


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ryan1203

引用:

原帖由 claywatts 於 2011-4-22 11:32 PM 發表
1) Find the remainder when 7^7000 + 7^700 + 7^70 +7 is divisible by 8


下面仲有兩題
希望有人答到我

1. x^(2n+1) + 1 = (x + 1)(x^2n - x^(2n-1) + x^(2n-2) + ... + x^2 - x + 1) for any positive integers n.
Put x = 7, we have 7^(2n+1) + 1 = 8M, where M is positive integer.

Put n = 3499, 349, 34 respectively, we have 7^6999 + 1 = 8P, 7^699 + 1 = 8Q, 7^69 + 1 = 8R,
where P, Q, R are positive integers.

Hence 7^7000 + 7^700 + 7^70 +7
= 7(7^6999 + 7^699 + 7^69 + 1)
= 7(8P + 8Q + 8R - 2)
= 8(7P + 7Q + 7R) - 14
= 8(7P + 7Q + 7R - 2) + 2
So the remainder is 2.

[ 本帖最後由 ryan1203 於 2011-4-23 02:51 PM 編輯 ]

dice

1. k = -9
2. k = 16

claywatts

factorize x^3 + 2y^3 - 3xy^2

charityho

f(x) = x^7000 + x^700 + x^70 +x
f(x) = (x+1)Q(x)+k        --> f(7) = 8Q(7) + k
f(-1) = k
    k = ???

charityho

for Q2
48   =  2*4*6
f(x) = (x+2)(x+4)(x+6)
so the root is -2 , -4 and -6