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storchak [24]
3 years ago
13

B=3x+9xy solv for x

Mathematics
2 answers:
IrinaVladis [17]3 years ago
8 0
The answer is  B = 12 x
Aliun [14]3 years ago
3 0

Answer:

x= \frac{B}{3(1 + 3y)}

or

x =  \frac{B}{(3 + 9y)}

Step-by-step explanation:

The given equation is B = 3x + 9xy

We have to solve for x.

B = 3x + 9xy

Let's factor the above equation.

B = 3x(1 + 3y)

We need to solve for x.

So, divide both sides by (1 + 3y), we get

\frac{B}{(1+ 3y)} = \frac{3x(1+3y)}{(1 +3y)}

\frac{B}{(1 + 3y)} = 3x

This can be written as

3x = \frac{B}{(1 + 3y)}

Now dividing both sides by 3, we get

x= \frac{B}{3(1 + 3y)}

or

x =  \frac{B}{(3 + 9y)}

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Tìm số dư trong phép chia 3^{2020} chia cho 13
djverab [1.8K]

Your question translates to computing 3^{2020} \pmod {13}.

Recall Euler's theorem: if \gcd(a,n)=1 (that is, a and n are relatively prime), then a^{\varphi(n)}\equiv1\pmod n, where \varphi(n) denotes Euler's totient function, which counts the number of positive integers relatively prime to n.

Since 13 is prime, we have \phi(13)=12. Then by Euler's theorem,

3^{12} \equiv 1 \pmod{13}

Now, observe that 2020 = 168×12 + 4, so that

3^{2020} \equiv 3^{168\times12+4} \equiv \left(3^{12}\right)^{168} \times 3^4 \equiv 1^{168} \times 3^4 \equiv 3^4 \pmod{13}

and since 3⁴ = 81 = 6×13 + 3, we end up with

3^{2020} \equiv 81 \equiv 3 \pmod{13}

so the remainder upon dividing 3²⁰²⁰ by 13 is 3.

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