100 Day Challenge 2020

Identity Boot Camp

Good day, and welcome to Identity Boot Camp!

  • Spoiler: No, this isn't going to be a page about finding yourself.

The term "identities" is used to mean several different things in mathematics. However, for the sake of today's problem and this introduction, we're going to define an identity as "an equation that is true no matter what values you substitute in for the variables used in it."

For example:

The Difference of Squares Identity
(a+b)(ab)=a2b2(a+b)(a-b)=a^2-b^2 For any two real numbers aa and b,b, the product of their sum and their difference is equal to the first of the two values squared minus the second of the two values squared.

Try testing any two real numbers into this equation for aa and b,b, and you'll see that it always holds true no matter what two numbers you choose. For example, (50+3)(503)=25009.(50+3)(50-3)=2500-9. You can use pen and paper or a calculator to check it if you want.

Identities can make solving a problem or simplifying out an algebraic description a lot easier, and today's challenge is no exception. In particular, the rest of this introduction defines and applies a set of relevant identities known as the linearity of averages.

Here's a problem to get things started. Try to solve it right now before we go through the solution analysis.

If, while solving this problem, you wrote out the statement "the average of 7x7x and 11x11x is 9090" algebraically, you probably realized that a lot of the algebra simplifies away: 7x+11x2=90\frac{7x + 11x}{2}=90 simplifies to 18x2=90,\frac{18x}{2}=90, which again simplifies to 9x=90.9x=90. Essentially, you just made use of this algebraic fact:

The average of 7a7a and 11a11a will always equal 9a,9a, no matter what aa is. Written out algebraically, this corresponds to the identity 7a+11a2=9a.\frac{7a + 11a}{2} = 9a.

Note that the equation 7a+11a2=9a\frac{7a + 11a}{2} = 9a doesn't give you any new information about the variable aa because it's true for all values of a.a.

Additionally, this single claim about one average can be generalized into a much broader observation about averages that's important enough to have a name: "the linearity of averages."

The linearity of averages is a combination of these two identities:

If you multiply any two numbers xx and yy each by the same number, a,a, then the average of the two new numbers is equal to aa times the average of the two original numbers:
(a×x)+(a×y)2=a×x+y2.\frac{(a\times x)+(a\times y)}{2} = a \times \frac{x+y}{2}. Additionally, if you add the same value, k,k, to each of two numbers, then the average of the two new numbers will be increased by kk compared to the average of the two original numbers: (x+k)+(y+k)2=x+y2+k.\frac{(x + k)+(y + k)}{2} = \frac{x+y}{2} + k.

This also makes sense if you think about it on a number line. If you're counting out on a number line by a’s,9aa\text{'s}, 9a will always be right in the middle between 7a7a and 11a.11a.

And this logic can be generalized to other numbers and larger averages. For example, the average of 2b,17b,2b, 17b, and 20b20b will be 2+17+203b=13b.\frac{2+17+20}{3} b = 13b.

So, congratulations! You probably haven't found yourself yet, but you have at least managed to find a useful, mathematical truth.

Today's Challenge

qq is equal to the product of 123123 and 456.456.

What is the value of r+s+t?r+s+t?


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