A number can be divided by 12, 15 and 18 at the same time. The minimum number is () and the prime factor is () RT, I've been talking about this for a long time,

A number can be divided by 12, 15 and 18 at the same time. The minimum number is () and the prime factor is () RT, I've been talking about this for a long time,

A number can be divided by 12, 15 and 18 at the same time. The minimum number is (180). The prime factor of decomposing the number is (180 = 2x2x3x5)
The least common multiple of 12, 15 and 18 is 180
The minimum number is (180), and the prime factor is (180 = 2x2x3x3x5)
180=2*2*3*3*5
A number can be divided by 12, 15 and 18 at the same time. The minimum number is (180). The prime factor of decomposing the number is (2 × 2 × 3 × 3 × 5)
A number can be divided by 12, 15 and 18 at the same time. The minimum number is (180). The prime factor of decomposing the number is (180 = 2 × 2 × 3 × 3 × 5)
Because the least common multiple of 12, 15 and 18 is 180
180=3×3×2×2×5
This integer is divided by 2 × 3, 180, so it can be divided by the smallest factor (15 × 2, 180)
Six pop cans, nine drink bottles, the price of each is the same, a total of 1.5 yuan. How much is each?
Use the slightly complex equation to solve. Set up the equivalent relation
Let's set each x yuan
Total price of pop can + total price of beverage bottle = 1.5 yuan
6X+9X=1.5
15X=1.5
X=1.5÷15
X=0.1
RMB 0.1 each
Do not understand can ask, help please adopt, thank you!
It is known that the sum of the first n terms of the equal ratio sequence {an} is Sn, a4-a1 = 78, S3 = 39, let BN = log3an, then the sum of the first 10 terms of the sequence {BN} is ()
A. log371B. 692C. 50D. 55
Let the common ratio of the equal ratio sequence {an} be Q. from a4-a1 = 78, S3 = 39, we can get a1q3-a1 = 78a1 + a1q + a1q2 = 39. By comparing the two formulas, we can get: Q-1 = 2, that is, q = 3. A1 (33-1) = 78, then A1 = 3. An = a1qn-1 = 3.3n-1 = 3N. BN = log3an = log33n = n +...
A number is not only a multiple of 3 and 4, but also a factor of 12. What is the number
It's 12 itself
Twelve
A number is not only a factor of 12, but also a multiple of 4
6 cans, 9 bottles, each price is the same, a total of 1.5 yuan
1.5 (6 + 9) = 1.5 (15) = 0.1 (yuan); answer: each 0.1 yuan
If the positive proportional sequence {an} satisfies A2 · A4 = 1, S3 = 13, BN = log3an, then the sum of the first 10 terms of the sequence {BN} is?
(1 / Q) ^ 2 + 1 / Q + 1 = 13?
Let the common ratio be q, from A2 · A4 = 1 to A3 = 1, from S3 = 13
(1 / Q) ^ 2 + 1 / Q + 1 = 13,
A1 = 1 / 9, an = (1 / 3) ^ (n + 1). BN = log3an = - n-1, this is the arithmetic sequence, the sum of the first 10 terms can be calculated~
a1=(1/q)^2
a2=1/q
a3=1
Because it is an equal ratio sequence, let the common ratio be Q. a3=1
a2=1／q
12 / 3 = 4, 3 and 4 are factors of 12, 12 is a multiple of 3 and 4, is that right
Yes
6 cans, 9 bottles, each price is the same, a total of 1.5 yuan
1.5 (6 + 9) = 1.5 (15) = 0.1 (yuan); answer: each 0.1 yuan
It is known that the sequence {an} is an equal ratio sequence, where A3 = 1 and A4, A5 + 1 and A6 are equal difference sequences. The first n terms of the sequence {an / BN} and Sn = (n-1) 2 ^ (n-2) + 1
(1) Find the general formula of sequence {an}, {BN}
(2) Let the sum of the first n terms of the sequence {BN} be TN. if t 3n-t n ≥ t holds for all positive integers n, the value range of real number T is obtained
(1)
If A4, A5 + 1 and A6 are equal difference sequence, then 2 (A5 + 1) = A4 + A6
A 4 = a 3q, a 5 = a 3Q & # 178; a 6 = a 3Q & # 179; a 3 = 1 is substituted and sorted out
q³-2q²+q-2=0
q²(q-2)+(q-2)=0
(q²+1)(q-2)=0
Q & # 178; + 1 is always positive. If the equation holds, only q = 2
a1=a3/q²=1/2²=1/4
an=(1/4)×2^(n-1)=2^(n-3)
The general formula of sequence {an} is an = 2 ^ (n-3)
S1=(1-1)×2^(1-2) +1=1 a1/b1=1 b1=a1=1/4
an/bn=Sn-Sn-1=(n-1)×2^(n-2)+1-(n-2)×2^(n-3)-1=n×2^(n-3)
bn=an/[n×2^(n-3)]=2^(n-3)/[n×2^(n-3)]=1/n
When n = 1, B1 = 1 / 4, not satisfied
The general formula of sequence {BN} is
bn=1/4 n=1
1/n n≥2
[T3(n+3)-T(n+1)]-(T3n-Tn)
=[1/4+1/2+1/3+...+1/(3n)+1/(3n+1)+1/(3n+2)+1/(3n+3)]-[1/4+1/2+1/3+...+1/n+1/(n+1)]
-[1/4+1/2+1/3+...+1/(3n)]+(1+1/2+1/3+...+1/n)
=1/(3n+1)+1/(3n+2)+1/(3n+3)-1/(n+1)
>1/(3n+3)+1/(3n+3)+1/(3n+3)-1/(n+1)
=1/(n+1)-1/(n+1)=0
T3(n+3)-T(n+1)>T3n-Tn
That is, with the increase of N, t 3N TN increases monotonically. When n = 1, t 3N TN reaches the minimum
T3-T1=(1/4+1/2+1/3)-(1/4)=1/2+1/3=5/6
Let t 3N - t n ≥ t hold for all positive integers n, as long as t ≤ 5 / 6
So 12 is a multiple and 4 is a factor______ (judge right or wrong)
Because 12 △ 4 = 3, 12 is a multiple of 4, and 4 is a factor of 12. The factor and multiple can not exist alone, so this question is wrong