Category Archives: Grades 9-12

Difference of 3-Digit Numbers with Reversed Digits

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In the previous post, we proved a theorem that the sum of a 2-digit number with reversed digit is divisible by 11. In this post, we will learn another theorem about numbers with reversed digits. This time, we will explore the difference between two 3-digit numbers with reversed digits.

Let’s have the following examples.

231 – 132 = 99

512 – 215 = 297

741 – 147 = 594

543 – 345 = 198

Before continuing reading, try other numbers and see if you can find a pattern.

You have probably observed that all the differences of a 3-digit number and a number formed by reversing its digit is a multiple by 99. Now, is this true for all possible cases? You may want to prove this on your own before continuing below. The proof on the 2-digit number with reversed digits is would be a good hint.  Continue reading →

The Sum and Product of Roots Theorem

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From the quadratic formula, we know that the numbers $latex r_1$ and $latex r_2$ are the roots of the quadratic equation $latex ax^2 + bx +c =0$ where $latex a \neq 0$ if and only if

$latex r_1 + r_2 = -\frac{b}{a}$

and

$latex r_1r_2 = \frac{c}{a}$.  Continue reading →

Sample Proof on Triangle Congruence Part 3

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Recall that for real numbers a, b, and c are real numbers, if $latex a = b$ and $latex b = c$, then $latex a = c$. This is called Transitive Property of Equality. This is also the same with congruence. If $latex A$, $latex B$, and $latex C$ are polygons, and if $latex A$ is congruent to $latex B$, and $latex B$ is congruent $latex C$, then $latex A$ is congruent to $latex C$. This is called the Transitive Property of Congruence. We will use this to prove the following problem.

Given: $latex \overline{BE} \cong \overline{DC}$ and $latex \overline{BD} \cong \overline{CA}$.

Prove: $latex \triangle DBE \cong \triangle CAB$.

Proof

It is given that $latex \overline{BE} \cong \overline{DC}$.

triangle congruence

Now, by reflexive property, that is a segment is congruent to itself, $latex \overline{BD} \cong \overline{BD}$.  Continue reading →