1. Find the difference between the largest number and the smallest prime number
from 120 to 190.
2. For integers that lie between 1 to 10 (excludinging 10 itself),
subtract the sum of all the prime numbers from the sum of all the composite
numbers, find the difference.
3.The fourth multiple of 9 is divided by the second multiple of 6.
Find the quotient.
4. Express each of 45 and 120 as a product of prime factors, Hence, find the
H.C.F. and L.C.M. of 45 and 120.
要有題解同橫式。
maths!!
2007-08-16 5:19 pm
回答 (2)
2007-08-16 5:27 pm
✔ 最佳答案
1. largest number=190, smallest prime number=127, difference=190-127=632. sum of all prime numbers=2+3+5+7=17, sum of all the composite
numbers=4+6+8+9=27, difference=10
3. 9*4/6*2=36/12=3
4. 45=3*3*5, 120=3*2*2*2*5
H.C.F.=3*5=15
L.C.M.=3*3*5*2*2*2=10*4*9=10*36=360
參考: my brain
2007-08-16 6:22 pm
1.
largest prime no. = 181 & smallest prime no. =127
difference = 181-127 = 54
2.
sum of prime nos. = 2+3+5+7 = 17
sum of all composite nos. = 4+6+8+9 = 27 (1 is not a prime no. nor a composite no.)
difference = 27 - 17 = 10
3.
4*9 / 2*6
= 36 /12
= 3
4.
45 = 3*3*5
120 = 2*2*2*3*5
HCF = 3*5 = 15
LCM = 2*2*2*3*3*5 = 360
largest prime no. = 181 & smallest prime no. =127
difference = 181-127 = 54
2.
sum of prime nos. = 2+3+5+7 = 17
sum of all composite nos. = 4+6+8+9 = 27 (1 is not a prime no. nor a composite no.)
difference = 27 - 17 = 10
3.
4*9 / 2*6
= 36 /12
= 3
4.
45 = 3*3*5
120 = 2*2*2*3*5
HCF = 3*5 = 15
LCM = 2*2*2*3*3*5 = 360
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