Math question help 2.0


  • M0★ M1 M2 M3 M4 M5

    The lines 3x+4y=5 and kx+12y=13 are perpendicular. Compute k.
    
    
    What is the value of 〖2-(-2)〗^(-2). 
    
    If the prime factorization of N! is 2^19∙3^9∙5^4∙7^3∙11^2∙13∙17∙19, compute N.
    
    In right triangle LUV with right angle U, LU = 12 and LV – UV = 9. Compute LV. Express your answer as a fraction in simplest form. 
    
    Compute the value of (2016!+2015!)/(2016!-2015!). Express your answer as a fraction in simplest form. 
    

    Question: how do you solve these questions? Is Prof. Loh going to teach similar lessons?


  • ADMIN M0★ M1 M5

    @The-Blade-Dancer Hi there! 🙂

    The first question is covered in the Module 4 Day 13 lesson.

    The second question has to do with exponents, covered in Module 4 Day 7, but it's not as hard as it looks. You can simplify the \(2 - (-2)\) as \(4\) and then just take this to the power \(-2.\) You get \(4^{-2} = \frac{1}{4^2} = \frac{1}{16}.\)

    The third question is about factorials, which are covered a bit in Module 3 Day 1, Module 3 Day 5, and Module 5 Day 6. The largest prime in that expression is \(19,\) so we know that \(N\) must be at least \(19,\) but it can't be larger than \(22,\) since \(23\) is a prime and isn't included in the factorization. You just need to try \(N = 19, N=20, N=21, N=22\) and see which of their prime factorizations matches. Looking at the power of \(7\) would help to identify if \(N\) is smaller or larger than \(21.\) Looking at the power of \(11\) helps with seeing if \(N\) is smaller or equal to \(22.\)

    For the fourth question, you can use the Pythagorean Theorem to solve for the side length, \(x,\) of the short leg. Since the hypotenuse is in terms of the short leg (it's \(9\) larger than the short leg), you only have one variable, so you can solve your equation. Pythagorean Theorem is covered in Module 0 Day 1 and Module 2 Day 1, among other lessons.

    For the last question, you can factor out \(2015!\) from the numerator and \(2015!\) from the denominator. Use the fact that \(2016! = 2016 \times 2015!.\)

    🙂


  • M0★ M1 M2 M3 M4 M5

    @The-Blade-Dancer said in Math question help 2.0:

    2^19∙3^9∙5^4∙7^3∙11^2∙13∙17∙19

    Gotcha, this helped a LOT. I couldn't even read the explanations they included before 😆

    But how does the prime factorization thing work? Isn't 2 to the power of 19 a lot larger than 19, 20, 21 , or 22?


  • ADMIN M0★ M1 M5

    @The-Blade-Dancer The factor of \(2^{19}\) comes from all the even numbers in the factorial \(N!,\) like \(2, 4, 6, 8, 12, 14, \ldots\) etc., in addition to extra \(2\)'s that come from the powers of \(2\) like \(4, 8, 16.\)

    Actually, though, you know that \(N\) must equal \(22,\) since we have two \(11\)'s (from the fact that \(11^2\) is a factor). One is from \(11,\) and the other is from \(22.\) We know that \(N\) can't be larger than \(22,\) so \(\boxed{N=22}.\)

    And then \(N!\) (N factorial) would be \(22! = 1 \times \textcolor{red}{2} \times 3 \times \textcolor{red}{4} \times 5 \times \textcolor{red}{6} \times \textcolor{blue}{7} \times \textcolor{red}{8} \times 9 \times \textcolor{red}{10} \times \textcolor{yellow}{11} \times \textcolor{red}{12} \times 13 \times \textcolor{purple}{14} \times 15 \times \textcolor{red}{16} \times 17 \times \textcolor{red}{18} \times 19 \times \textcolor{red}{20} \times \textcolor{blue}{21} \times \textcolor{orange}{22}.\)

    I tried to color-code so that all of the \(\textcolor{red}{\text{red}}\) numbers are even and contribute factors of \(2\) (this is where you get the \(2^{19}\)), the \(\textcolor{blue}{\text{blue}}\) numbers contribute factors of \(7\), and the yellow numbers contribute factors of \(11.\) Since \(\textcolor{purple}{14}\) is a multiple of both \(2\) and \(7,\) I made it purple. And since \(22\) is a multiple of both \(11\) and \(2,\) I made it orange.
    🙂


Log in to reply