Math question help 2.0

• 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?

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 $$2016! = 2016 \times 2015!.$$

• 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?

• @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.