送10分!!!!!! 數學題(easy)奧maths3

(1) 3^2111 = 3^(4x527+3) = 3^(4x527)x3^3 = (3^4)^527 x 3^3 = 81^527 x 27
Since the unit digit of 81^527 is 1, the unit digit of 81^527 x 27 is 7.
Therefore, the remainder of 3^2111÷10 is 7.

(2) 2^1234 x 7^3312 x 25^4564 = 2^1234 x 7^3312 x (5^2)^4564
= 2^1234 x 7^3312 x 5^9128
= (2x5)^1234 x (7x5)^3312 x 5^4582
= 10^1234 x 35^3312 x 5^4582
Since the number is a multiple of 10^1234, its unit digit is 0.

(3) 1÷(2÷3)÷(4÷5÷6)÷(2÷3÷2)÷(1/2) = 1÷(2/3)÷(2/15)÷(1/3)÷(1/2)
= 1÷[(2/3)x(2/15)x(1/3)x(1/2)]
= 1÷(2/135)
= 135/2

(1)
3^n 個位數的循環是 3 -> 9 -> 7 -> 1，循環周期=4，
2111 (mod 4) = 3
所以，2^2111的個位數是7, 而3^2111÷10的餘數是7。


(2)
2^n 個位數的循環是 2 -> 4 -> 8 -> 6，循環周期=4
1234 (mod 4) = 2
所以，2^1234的個位數是4;

7^n 個位數的循環是 7 -> 9 -> 3 -> 1，循環周期=4
3312 (mod 4) = 0
所以，7^3312的個位數是1;

25^n 個位數永遠都是5
所以，25^4564的個位數是5;

綜合結果，2^1234 x 7^3312 x 25^4564的個位數=4*1*5的個位數 = 0。


(3) 1÷(2÷3)÷(4÷5÷6)÷(2÷3÷2)÷1/2=
= 1÷2x3÷4x5x6÷2x3x2x2
= 135/2

1. 餘數是3
2. 個位是0
3. 135/2