Lizzie came up with a divisibility test for a certain number $m \neq 1$:
Break a positive integer $n$ into two-digit chunks, starting from the ones place. (For example, the number $354764$ would break into the two-digit chunks $64, 47, 35.$)
Find the alternating sum of these two-digit numbers, by adding the first number, subtracting the second, adding the third, and so on. (In our example, this alternating sum would be $64-47+35=52.$)
Find $m,$ and show that this is indeed a divisibility test for $m$ (by showing that $n$ is divisible by $m$ if and only if the result of this process is divisible by $m$).

Given Lizzie's divisibility test, let's determine the number $ m $ and prove its validity.

To start, let’s define the given problem clearly with an example and then generalize it. 

Consider a positive integer $ n $ and break it into two-digit chunks starting from the ones place. For example, for $ n = 354764 $, the two-digit chunks are $ 64, 47, 35 $. The alternating sum of these chunks is $ 64 - 47 + 35 $.

We denote $ n $ as $ n = \sum_{i=0}^{k} a_i 100^i $, where $ a_i $ represents each two-digit chunk of $ n $ (if the highest chunk has less than two digits, it is padded with zeros). 

We want to find $ m $ such that $ n $ is divisible by $ m $ if and only if the alternating sum $ S = a_0 - a_1 + a_2 - a_3 + \cdots $ is divisible by $ m $.

Let's compute $ n \mod 101 $:

\[
n = \sum_{i=0}^{k} a_i 100^i
\]

First, observe the powers of 100 modulo 101:

\[
100 \equiv -1 \pmod{101}
\]

Thus:

\[
100^0 \equiv 1 \pmod{101}
\]
\[
100^1 \equiv -1 \pmod{101}
\]
\[
100^2 \equiv 1 \pmod{101}
\]
\[
100^3 \equiv -1 \pmod{101}
\]
\[
\text{and so on.}
\]

Using this pattern, we have:

\[
n \equiv a_0 (-1)^0 + a_1 (-1)^1 + a_2 (-1)^2 + a_3 (-1)^3 + \cdots \pmod{101}
\]
\[
n \equiv a_0 - a_1 + a_2 - a_3 + \cdots \pmod{101}
\]

This is exactly the alternating sum $ S $. Therefore, $ n \equiv S \pmod{101} $, meaning $ n $ is divisible by 101 if and only if $ S $ is divisible by 101.

Thus, the number $ m $ is 101. 

To conclude, the divisibility test proposed by Lizzie works because $ n $ is divisible by $ 101 $ if and only if the alternating sum of the two-digit chunks of $ n $ is divisible by 101.

Would you like details or have any questions? Here are 5 related questions you might want to explore:

1. How does this test work with smaller numbers, such as 3-digit or 4-digit integers?
2. Can this divisibility rule be extended to other bases, such as base 8 or base 16?
3. How does this test compare to traditional divisibility rules, like those for 3 or 9?
4. Can you derive a similar rule for divisibility by 11, using alternating sums?
5. What are some practical applications of this divisibility rule in number theory or cryptography?

**Tip:** When dealing with modular arithmetic, identifying patterns in the remainders of powers can simplify calculations and reveal interesting properties.

Explore Lizzie's divisibility test using alternating sums of two-digit chunks of a number. Understand how this test works, its proof using modular arithmetic, and its application in number theory. Suitable for advanced high school students and beyond.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation