F.1 代數, Please Help

1.a When a positive number is divided by 6, 9 and 15, the remainders are all 2. Find the smallest possible number.

The LCM of 6, 9, 15 = 90 , the smallest possible number = 90+2 = 92.

b. When a 3-digit number is divided by 8, the remainder is 2. When it is divided by 12, the remainder is 6. Find the largest and smallest possible value of the number.

Add 6 to the 3-digit number, then the remainder of the new number which is divided by 8 is 2+6 = 8 , i.e. the remainder = 0
the remainder of the new number which is divided by 12 is 6+6 = 12 ,i.e.= 0

The LCM of 8 and 12 = 24 ,

so the smallest required 3-digit number = 24*5 - 6 = 114
the largest required 3-digit number = 24*41 - 6 = 978



2. Write down a 2-digit number which is not prime but co-prime with 40, i.e. having no common factor with 40 except 1.

40 = 2*2*2*5,

many number can be answer : (not include the factor 2 and 5)

3*7 = 21 , 3*11 = 33 , 3*13 = 39 , 7*11 = 77 , etc.



3. Write down a 2-digit number which is exactly 3 factors

The number must be a complete square : only two answers :

25 (1 , 5 , 25)
49 (1 , 7 , 49)


4.If A is a prime number, what is the sum of all the factors of A ? Express your answers in terms of A.

The factors of a prime number A are 1 and itself , i.e. 1 and A

A + 1



5. List all the factors of 200.

200 = 2 * 2 * 2 * 5 * 5

= 2 ^ 3 * 5 ^ 2

Total ( 3 + 1 )( 2 + 1 ) = 4 * 3 = 12 factors :

2^0 * 5^0 = 1 ; 2^0 * 5^1 = 5 ; 2^0 * 5^2 = 25

2^1 * 5^0 = 2 ; 2^1 * 5^1 = 10 ; 2^1 * 5^2 = 50

2^2 * 5^0 = 4 ; 2^2 * 5^1 = 20 ; 2^2 * 5^2 = 100

2^3 * 5^0 = 8 ; 2^3 * 5^1 = 40 ; 2^3 * 5^2 = 200

Dear Mr. Leung,
Thank you for your help.

PM