+166 1.
Given that \(\frac{a}{25-a}+\frac{b}{65-b}+\frac{c}{60-c}=7\)
evaluate \(\frac{5}{25-a}+\frac{13}{65-b}+\frac{12}{60-c}\)
2.
What is the sum of all positive integers that have twice as many digits when written in base 2 as they have when written in base 3? Express your answer in base 10.
3.
Compute the multiplicative inverse of 201 modulo 299. Express your answer as an integer from 0 to 298.
4. Determine the number of integers x such that \(0\le x< 12\)
and x satisfies the following system of equations:
\(\begin{align*} &x-1\equiv 1-x\pmod {12},\\ &x-2\equiv 2-x\pmod{12}. \end{align*}\)
+981 Here is the full solution.
Just as guest said.
1.
\(\frac{a}{25-a}+\frac{b}{65-b}+\frac{c}{60-c}=\frac{25-25+a}{25-a}+\frac{65-65+b}{65-b}+\frac{60-60+c}{60-c}\)
\((\frac{25}{25-a}-1)+(\frac{65}{65-1}-1)+(\frac{60}{60-c}-1)=\frac{25}{25-a}+\frac{65}{65-b}+\frac{60}{60-c}-3\)
After factoring out the five, we have:
\(5(\frac{5}{25-a}+\frac{13}{65-b}+\frac{12}{60-c})-3=7\)
We can see that the value of what we are trying to find and its relantionship to the original expression.
Since the original expression is 7, and the expression we are trying to find is 3 less than 5 times 7,
We have:
\((7+3)\div5=2\)
Just the way guest did it!
I hope this helped,
Gavin.
I believe there's a simpler way for 1:
\(\frac{a}{25-a}+\frac{b}{65-b}+\frac{c}{60-c}=\frac{25-(25-a)}{25-a}+\frac{65-(65-b)}{65-b}+\frac{60-(60-c)}{60-c}=\\ (\frac{25}{25-a}-1)+(\frac{65}{65-b}-1)+(\frac{60}{60-c}-1)=\frac{25}{25-a}+\frac{65}{65-b}+\frac{60}{60-c}-3=\\ 5(\frac{5}{25-a}+\frac{13}{65-b}+\frac{12}{60-c})-3=7\\ \frac{5}{25-a}+\frac{13}{65-b}+\frac{12}{60-c}=x\Rightarrow5*x-3=7\Rightarrow5*x=10\Rightarrow x=2\)
1) a/(25 - a) + b/(65 - b) + c/(60 - c) = 7 This reduces to:
5/(a - 25) + 13/(b - 65) + 12/(c - 60) + 2 = 0 Add this to the 2nd equation in the question:
[5/(25 - a) + 13/(65 - b) + 12/(60 - c)] + [5/(a - 25) + 13/(b - 65) + 12/(c - 60)] + 2 = 2 - All terms cancel out leaving 2 as the answer.
+981 3.
The modular multiplicative inverse of an integer a modulo m is an integer b such that
\(ab \equiv 1 \pmod m\), It maybe noted \(a^{-1}\), where the fact that the inversion is m-modular is implicit.
The multiplicative inverse of a modulo m exists if and only if a and m are coprime (i.e., if gcd(a, m) = 1).
If the modular multiplicative inverse of a modulo m exists, the operation of division by a modulo m can be defined as multiplying by the inverse.
Zero has no modular multiplicative inverse.
The modular multiplicative inverse of a modulo m can be found with the Extended Euclidean algorithm.
To show this, let's look at this equation: ax + my = 1
This is linear diophantine equation with two unknowns, refer to Linear Diophantine equations.
Since one can be divided without reminder only by one, this equation has the solution only if \( {\rm gcd}(a,m)=1\). The solution can be found with the Extended Euclidean algorithm. The modulo operation on both parts of equation gives us \(ax = 1 \pmod m \)Thus, x is the modular multiplicative inverse of a modulo m.
The answer you seek is 180
I hope this helped,
gavin
