Engaging students: Reducing fractions to lowest terms

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Madison duPont. Her topic, from Pre-Algebra: reducing fractions to lowest terms.

How can this topic be used in your students’ future courses in mathematics or science?

Reducing fractions to lowest terms can be applied to future mathematics topics such as ratios and proportions, and scientific topics such as chemistry or physics. Ratios can be represented as fractions and are not typically reduced to lowest terms because they represent relationships of two subjects using numbers. Being able to reduce these ratios can help students better identify the underlying relationship and apply this relationship to other aspects of the math problem, such as problems using unit price or map scales. Proportions relate to the concept of reducing fractions to lowest terms when using cross-multiplication. Having both sides of the proportion reduced to lowest terms makes the cross-multiplication much easier to compute and derive a final reduced answer. Chemistry uses fractions reduced to lowest terms with topics, like stoichiometry, that use potentially small and large numbers in several ratios that are multiplied together to obtain a final converted and reduced answer. Physics often uses ratio-like formulas and problems that are applied to real-world scenarios, which typically require fractions reduced to lowest terms because answers like miles per one hour are the goal. All of these topics use concepts of reducing fractions to lowest terms to more easily accomplish problems using a series of fractional computations, or to get an answer that is in terms of a single unit or most reduced so that it makes sense to real-world application.

How does this topic extend what your students should have learned in previous courses?

This topic extends previously learned topics such as concepts of unique prime factorizations, greatest common divisor, manipulating fractions, and multiplication facts. The concept of unique prime factorizations greatly aids students in finding the greatest common divisor, which is used to find the greatest factor of the value of both the numerator and denominator. Next, manipulation of fractions is used to properly divide the numerator and denominator by the greatest common divisor. This process of dividing both parts of the fraction utilizes multiplication facts as well to determine what the answer to the division problem on both the top and bottom of the fraction would be. These previously learned concepts are all subtle and important applications when reducing fractions to lowest terms.

How can technology (YouTube, Khan Academy [khanacademy.org], Vi Hart, Geometers Sketchpad, graphing calculators, etc.) be used to effectively engage students with this topic? Note: It’s not enough to say “such-and-such is a great website”; you need to explain in some detail why it’s a great website.

This video reminded me of many students that I have tutored or encountered in classrooms that were determined that a calculator was all they needed when doing math. Applied to reducing fractions to lowest terms, this video is extremely relevant in displaying that technology cannot be the only source of intelligence when thinking mathematically. Reducing fractions with extremely large numbers or numbers that do not have well-known factors can seem exhausting or impossible. Punching several factors of the numerator and denominator into a calculator attempting to reduce numbers with each common factor, and then not being sure of whether the fraction appearing on their screen is truly in the most reduced form surely indicates the technology is not the only way of solving the problem. Many students hop on a procedural escalator when beginning varying types of problems (in addition to reducing fractions to lowest terms) using memorized steps, punching calculator buttons, feeling comfortable, until suddenly—there is a horribly unattractive fraction halting their progress. This is when using mathematical problem solving skills such as reducing the numerator and denominator by the greatest common divisor or checking to see that the numerator and denominator are relatively prime becomes pertinent. Using these conceptual skills can save someone that is stuck waiting for a calculator to do the work for them, or that has given up on finishing a problem because it seems impossible or difficult, from thinking they are incapable of working out a problem efficiently and successfully. This video highlights the importance of being capable of knowing when it is time to take the effort to climb the stairs to reach your destination.

References:

“Stuck on an Escalator” Video link:

https://www.youtube.com/watch?v=VrSUe_m19FY

found via Google video search

Lessons from teaching gifted elementary school students: Index (updated)

I’m doing something that I should have done a long time ago: collect past series of posts into a single, easy-to-reference post. The following posts formed my series on various lessons I’ve learned while trying to answer the questions posed by gifted elementary school students. (This is updated from my previous index.)

Part 1: A surprising pattern in some consecutive perfect squares.

Part 2: Calculating 2 to a very large exponent.

Part 3a: Calculating 2 to an even larger exponent.

Part 3b: An analysis of just how large this number actually is.

Part 4a: The chance of winning at BINGO in only four turns.

Part 4b: Pedagogical thoughts on one step of the calculation.

Part 4c: A complicated follow-up question.

Part 5a: Exponentiation is multiplication as multiplication is to addition. So, multiplication is to addition as addition is to what? (I offered the answer of incrementation, but it was rejected: addition requires two inputs, while incrementation only requires one.)

Part 5b: Why there is no binary operation that completes the above analogy.

Part 5c: Knuth’s up-arrow notation for writing very big numbers.

Part 5d: Graham’s number, reputed to be the largest number ever to appear in a mathematical proof.

Part 6a: Calculating $(255/256)^x$.

Part 6b: Solving $(255/256)^x = 1/2$ without a calculator.

Part 7a: Estimating the size of a 1000-pound hailstone.

Part 7b: Estimating the size a 1000-pound hailstone.

Part 8a: Statement of an usually triangle summing problem.

Part 8b: Solution using binomial coefficients.

Part 8c: Rearranging the series.

Part 8d: Reindexing to further rearrange the series.

Part 8e: Rewriting using binomial coefficients again.

Part 8f: Finally obtaining the numerical answer.

Part 8g: Extracting the square root of the answer by hand.

Pizza Hut Pi Day Challenge (Part 8)

On March 14, 2016, Pizza Hut held a online math competition in honor of Pi Day, offering three questions posed by Princeton mathematician John H. Conway. As luck would have it, years ago, I had actually heard of the first question before from a colleague who had heard it from Conway himself:

I’m thinking of a ten-digit integer whose digits are all distinct. It happens that the number formed by the first n of them is divisible by n for each n from 1 to 10. What is my number?

I really like this problem because it’s looks really tough but only requires knowledge of elementary-school arithmetic. So far in this series, I described why the solution must be one of the following 10 numbers:

1 , 4 7 2 , 5 8 9 , 6 3 0,

7 , 4 1 2 , 5 8 9 , 6 3 0,

1 , 8 9 6 , 5 4 3 , 2 7 0,

9 , 8 1 6 , 5 4 3 , 2 7 0,

7 , 8 9 6 , 5 4 3 , 2 1 0,

9 , 8 7 6 , 5 4 3 , 2 1 0,

1 , 8 3 6 , 5 4 7 , 2 9 0,

3 , 8 1 6 , 5 4 7 , 2 9 0,

1 , 8 9 6 , 5 4 7 , 2 3 0,

9 , 8 1 6 , 5 4 7 , 2 3 0.

Up until now, I have used the divisibility rules to ensure that the property works for n = 1, 2, 3, 4, 5, 6, 8, 9, 10. But I haven’t used n = 7 yet.

Step 10. The number formed by the first seven digits must be a multiple of 7. There is a very complicated divisibility rule for checking to see if a number is a multiple of 7. However, at this point, it’s easiest to just divide by 7 and see what happens.

$1 , 4 7 2 , 5 8 9 / 7 = 210,369.857\dots$ : not a multiple of 7.

$7 , 4 1 2 , 5 8 9 / 7 = 1,058,941.285\dots$ : not a multiple of 7.

$1 , 8 9 6 , 5 4 3 / 7 = 270,934.714\dots$ : not a multiple of 7.

$9 , 8 1 6 , 5 4 3 / 7 = 1,402,363.285\dots$ : not a multiple of 7.

$7 , 8 9 6 , 5 4 3 / 7 = 1,128,077.571\dots$ : not a multiple of 7.

$9 , 8 7 6 , 5 4 3 / 7 = 1,410,934.714\dots$ : not a multiple of 7.

$1 , 8 3 6 , 5 4 7 = 262,363.857\dots$ : not a multiple of 7.

$3 , 8 1 6 , 5 4 7 / 7 = 545,221$ : a multiple of 7!!!

$1 , 8 9 6 , 5 4 7 / 7 = 270,935.285\dots$ : not a multiple of 7.

$9 , 8 1 6 , 5 4 7 / 7 = 1,402,363.857\dots$ : not a multiple of 7.

So, by inspection, only one of these works, yielding the answer to the puzzle:

3,816,547,290.

Pizza Hut Pi Day Challenge (Part 7)

I’m thinking of a ten-digit integer whose digits are all distinct. It happens that the number formed by the first n of them is divisible by n for each n from 1 to 10. What is my number?

I really like this problem because it’s looks really tough but only requires knowledge of elementary-school arithmetic. So far in this series, I described why the solution must have one of the following four forms:

O , 4 O 2 , 5 8 1 , 6 O 0,

O , 4 O 2 , 5 8 9 , 6 O 0,

O , 8 O 6 , 5 4 3 , 2 O 0,

O , 8 O 6 , 5 4 7 , 2 O 0.

where O represents the remaining three odd digts. (The last digit is 0 and not an odd number.) There are 24 possible answers left.

Step 9. The number formed by the first three digits must be a multiple of 3. By the divisibility rules, this means that the sum of the first three digits must be a multiple of 3.

For the first form, that means that O + 4 + O must be a multiple of 3, where O is chosen from the remaining odd digits, which are 3, 7, and 9. We can directly test this to see that it’s impossible:

3 + 4 + 7 = 14, not a multiple of 3.

3 + 4 + 9 = 16, not a multiple of 3.

7 + 4 + 9 = 20, not a multiple of 3.

For the second form, that means that O + 4 + O must be a multiple of 3, where O is chosen from the remaining odd digits, which are 1, 3, and 7. We can directly test this:

1 + 4 + 3 = 8, not a multiple of 3.

1 + 4 + 7 = 12, a multiple of 3.

3 + 4 + 7 = 14, not a multiple of 3.

Therefore, for the second form, the first three digits could be either 147 or 714.

For the third form, that means that O + 8 + O must be a multiple of 3, where O is chosen from the remaining odd digits, which are 1, 7, and 9. We can directly test this to see that it’s impossible:

1 + 8 + 7 = 16, not a multiple of 3.

1 + 8 + 9 = 18, a multiple of 3.

7 + 8 + 9 = 24, a multiple of 3.

Therefore, for the third form, the first three digits could be either 189, 981, 789, or 987.

For the fourth form, that means that O + 8 + O must be a multiple of 3, where O is chosen from the remaining odd digits, which are 1, 3, and 9. We can directly test this to see that it’s impossible:

1 + 8 + 3 = 12, a multiple of 3.

1 + 8 + 9 = 18, a multiple of 3.

3 + 8 + 9 = 20, not a multiple of 3.

Therefore, for the fourth form, the first three digits could be 183, 381, 189, or 981.

1 , 4 7 2 , 5 8 9 , 6 O 0,

7 , 4 1 2 , 5 8 9 , 6 O 0,

1 , 8 9 6 , 5 4 3 , 2 O 0,

9 , 8 1 6 , 5 4 3 , 2 O 0,

7 , 8 9 6 , 5 4 3 , 2 O 0,

9 , 8 7 6 , 5 4 3 , 2 O 0,

1 , 8 3 6 , 5 4 7 , 2 O 0,

3 , 8 1 6 , 5 4 7 , 2 O 0,

1 , 8 9 6 , 5 4 7 , 2 O 0,

9 , 8 1 6 , 5 4 7 , 2 O 0.

Indeed, for each of these, there is only one odd digit left, which means we automatically know what it has to be for each of these 10 answers by process of elimination:

1 , 4 7 2 , 5 8 9 , 6 3 0,

7 , 4 1 2 , 5 8 9 , 6 3 0,

1 , 8 9 6 , 5 4 3 , 2 7 0,

9 , 8 1 6 , 5 4 3 , 2 7 0,

7 , 8 9 6 , 5 4 3 , 2 1 0,

9 , 8 7 6 , 5 4 3 , 2 1 0,

1 , 8 3 6 , 5 4 7 , 2 9 0,

3 , 8 1 6 , 5 4 7 , 2 9 0,

1 , 8 9 6 , 5 4 7 , 2 3 0,

9 , 8 1 6 , 5 4 7 , 2 3 0.

So we’re down to 10 possible answers left.

In tomorrow’s post, I’ll finally find the answer.

Pizza Hut Pi Day Challenge (Part 6)

I’m thinking of a ten-digit integer whose digits are all distinct. It happens that the number formed by the first n of them is divisible by n for each n from 1 to 10. What is my number?

O , 4 O 2 , 5 8 O , 6 O 0,

O , 6 O 2 , 5 8 O , 4 O 0,

O , 2 O 6 , 5 4 O , 8 O 0.

O , 8 O 6 , 5 4 O , 2 O 0.

where O is one of 1, 3, 7, and 9. (The last digit is 0 and not an odd number.) There are 96 possible answers left.

Step 8. The number formed by the first eight digits must be a multiple of 8. By the divisibility rules, this means that the number formed by the sixth, seventh, and eighth digits must be a multiple of 8.

For the first form, that means that $8O6$ must be a multiple of 8. We can directly test this:

$816/8 = 102$ : a multiple of 8.

$836/8 = 104.5$ : not a multiple of 8.

$876/8 = 109.5$ : not a multiple of 8.

$896/8 = 112$ : a multiple of 8.

For the second form, that means that $8O4$ must be a multiple of 8. This is impossible. Let $O = 2n+1$ . Then

$8O4 = 800 + 10 \times O + 4$

$= 800 + 10(2n+1) + 4$

$= 800 + 20n + 14$

$= 2(400 + 10n + 7)$ .

We see that $7$ is odd, and therefore $400 + 10n + 7$ is not a multiple of 2. Therefore, $2(400 + 10n + 7)$ is not a multiple of 4 (let alone 8).

For the third form, that means that $4O8$ must be a multiple of 8. This is also impossible. Let $O = 2n+1$ . Then

$4O8 = 400 + 10 \times O + 8$

$= 400 + 10(2n+1) + 8$

$= 400 + 20n + 18$

$= 2(200 + 10n + 9)$ .

We see that $9$ is odd, and therefore $200 + 10n + 9$ is not a multiple of 2. Therefore, $2(200 + 10n + 9)$ is not a multiple of 4 (let alone 8).

For the fourth form, that means that $4O2$ must be a multiple of 8. We can directly test this:

$412/8 = 51.5$ : not a multiple of 8.

$432/8 = 54$ : a multiple of 8.

$472/8 = 59$ : a multiple of 8.

$492/8 = 61.5$ : not a multiple of 8.

In other words, we’re down to

O , 4 O 2 , 5 8 1 , 6 O 0,

O , 4 O 2 , 5 8 9 , 6 O 0,

O , 8 O 6 , 5 4 3 , 2 O 0,

O , 8 O 6 , 5 4 7 , 2 O 0.

For each of these, there are 3! = 6 ways of choosing the remaining odd digits. Since there are four forms, there are 4 x 6= 24 possible answers left.

In tomorrow’s post, I’ll cut this number down to 10.

Pizza Hut Pi Day Challenge (Part 5)

I’m thinking of a ten-digit integer whose digits are all distinct. It happens that the number formed by the first n of them is divisible by n for each n from 1 to 10. What is my number?

O , E O 2 , 5 8 O , E O 0,

O , E O 8 , 5 2 O , E O 0,

O , E O 4, 5 6 O , E O 0,

O , E O 6 , 5 4 O , E O 0.

where E is one of 2, 4, 6, and 8 (not already included) and O is one of 1, 3, 7, and 9. (The last digit is 0 and not an odd number.) There are 192 possible answers left.

Step 7. The number formed by the first four digits must be a multiple of 4. By the divisibility rules, this means that the number formed by the third and fourth digits must be a multiple of 4.

In other words, the two digit number $OE$ has to be a multiple of 4, where $O$ is odd and $E$ is even. Let $O = 2n+1$ and $E = 2k$ . Then

$OE = 10 \times O + E$

$= 10(2n+1) + 2k$

$= 20n+2k+10$

$= 2(10n+k+5)$

also has to be a multiple of 4, which means that $10n+k+5$ has to be a multiple of 2. Since $10n$ is already of a multiple of 2, that means that $k + 5$ must be a multiple of 2, or that $k$ must be odd.

In other words, $E = 2k$ , where $k$ is an odd number. Therefore, the fourth digit must be either 2 or 6. This eliminates two of the above forms, so we’re down to

O , E O 2 , 5 8 O , E O 0,

O , E O 6 , 5 4 O , E O 0.

At this point, I can include the remaining ways of choosing the even digits:

O , 4 O 2 , 5 8 O , 6 O 0,

O , 6 O 2 , 5 8 O , 4 O 0,

O , 2 O 6 , 5 4 O , 8 O 0.

O , 8 O 6 , 5 4 O , 2 O 0.

For each of these, there are 4! = 24 ways of choosing the remaining odd digits. Since there are four forms, there are 4 x x 24 = 96 possible answers left.

In tomorrow’s post, I’ll cut this number down to 24.

Pizza Hut Pi Day Challenge (Part 4)

I’m thinking of a ten-digit integer whose digits are all distinct. It happens that the number formed by the first n of them is divisible by n for each n from 1 to 10. What is my number?

I really like this problem because it’s looks really tough but only requires knowledge of elementary-school arithmetic. So far in this series, I described why the solution must have the form

O , E O E , 5 E O , E O 0,

where E is one of 2, 4, 6, and 8 and O is one of 1, 3, 7, and 9. (The last digit is 0 and not an odd number.) There are 576 possible answers left.

Step 6. The number formed by the first three digits must be a multiple of 3. Also, the number formed by the first six digits must be a multiple of 6, which means it’s also a multiple of 3. By the divisibility rules, the sum of the first three digits must be a multiple of 3, and the sum of the first six digits must be a multiple of 3. Therefore, the sum of the fourth, fifth, and sixth digits must be 3.

Stated another way, the fourth and sixth digits have to be different even numbers so that 5 plus the sum of these two even numbers is a multiple of 3. Some quick testing reveals that these two even numbers have to be either 4 and 6 or else 2 and 8 in some order:

5 + 2 + 4 = 11, not a multiple of 3.

5 + 2 + 6 = 13, not a multiple of 3.

5 + 2 + 8 = 15, a multiple of 3.

5 + 4 + 6 = 15, a multiple of 3.

5 + 4 + 8 = 17, a multiple of 3.

5 + 6 + 8 = 19, not a multiple of 3.

Therefore, the solution must have one of the following forms:

O , E O 2 , 5 8 O , E O 0,

O , E O 8 , 5 2 O , E O 0,

O , E O 4, 5 6 O , E O 0,

O , E O 6 , 5 4 O , E O 0.

For each of these there are 2! = 2 ways of choosing the remaining even digits and 4! = 24 ways of choosing the remaining odd digits. Since there are four forms, there are 4 x 2 x 24 = 192 possible answers left.

In tomorrow’s post, I’ll cut this number in half.

Pizza Hut Pi Day Challenge (Part 3)

I’m thinking of a ten-digit integer whose digits are all distinct. It happens that the number formed by the first n of them is divisible by n for each n from 1 to 10. What is my number?

I really like this problem because it’s looks really tough but only requires knowledge of elementary-school arithmetic. In yesterday’s post, I described why the solution must have the form

___ , ___ ___ ___ , 5 ___ ___ , ___ ___ 0,

where the blanks have to filled by using the numbers 1, 2, 3, 4, 6, 7, 8, and 9. There are 8 remaining digits, and so there are 8! = 40,320 possible answers left.

Step 5. The number formed by the first two digits has to be a multiple of 2. The number formed by the first four digits has to be a multiple of 4, which means it’s also a multiple of 2. The number formed by the first six digits has to be a multiple of 6, which means it’s also a multiple of 2. The number formed by the first eight digits has to be a multiple of 8, which means it’s also a multiple of 2. By the divisibility rules, that means that the second, fourth, sixth, and eighth digits are even numbers. So the ten-digit number must have the form

___ , E ___ E , 5 E ___ , E ___ 0,

where E is one of 2, 4, 6, and 8. Therefore, the remaining dashes must be odd numbers:

O , E O E , 5 E O , E O 0,

where O is one of 1, 3, 7, and 9. There are 4! = 24 ways of arranging the four remaining even digits and also 4! = 24 ways of arranging the remaining odd digits. Therefore, there are 24 x 24 = 576 possible answers left.

While it’s now tractable to find this number by brute force by listing and testing the 576 possible numbers, we can use the divisibility rules to reduce the list to only 192. I’ll continue this in tomorrow’s post.

Pizza Hut Pi Day Challenge (Part 2)

I’m thinking of a ten-digit integer whose digits are all distinct. It happens that the number formed by the first n of them is divisible by n for each n from 1 to 10. What is my number?

I really like this problem because it’s looks really tough but only requires knowledge of elementary-school arithmetic. To illustrate this idea, let me give an example that almost works: 9,632,581,470, a ten-digit number that uses each digit exactly once.

The first digit is 9, which is clearly a multiple of 1.
The first two digits form the number 96, which is clearly a multiple of 2: $96 = 2 \times 48$ .
The first three digits form the number 963, which is a multiple of 3: $963 = 3 \times 321$ .
The first four digits form the number 9632, which is a multiple of 4: $9632 = 4 \times 2408$ .
The first five digits form the number 96,325, which is a multiple of 5: $96,325 = 5 \times 19,265$ .
The first six digits form the number 963,258, which is a multiple of 6: $963,258 = 6 \times 160,523$ .
The first seven digits form the number 9,632,581, which is a multiple of 7: $9,632,581 = 7 \times 1,376,083$ .
Doesn’t work: $96,325,814 = 8 \times 12,040,726 + 6$ .
The first nine digits for the number 963,258,147, which is a multiple of 9: $963,258,147 = 9 \times 107,028,683$
The ten-digit number is a multiple of 10: $9,632,581,470 = 10 \times 963,258,147$ .

So, this number works for 9 of the 10 cases… but we need the number that works for all 10 cases.

Here’s the ten-digit number, where the blanks have to be filled in:

___ , ___ ___ ___ , ___ ___ ___ , ___ ___ ___.

Each of the blanks has to be a digit from 0 through 9, and each can only be used once. Therefore, there are 10! = 3,628,800 possible answers, but we’d like to whittle this down to a more tractable number.

Step 1. This ten-digit number has to be a multiple of 10. By the divisibility rules, that means that the last digit has to be 0. So the ten-digit number must have the form

___ , ___ ___ ___ , ___ ___ ___ , ___ ___ 0.

Each of the blanks has to be a digit from 1 through 9, and each can only be used once. Therefore, at this stage, there are 9! = 362,880 possible answers.

Step 2. The number formed by the first five digits has to be a multiple of 5. By the divisibility rules, that means that the fifth digit has to be either 5 or 0. However, the 0 has already been used, so the fifth digit must be 5. So the ten-digit number must have the form

___ , ___ ___ ___ , 5 ___ ___ , ___ ___ 0.

Each of the blanks has to be one of the 8 remaining digits, and each can only be used once. Therefore, at this stage, there are 8! = 40,320 possible answers.

Step 3. The number formed by the first nine digits has to be a multiple of 9. By the divisibility rules, that means that the sum of the first nine digits has to be a multiple of 9. However, this isn’t helpful: we know that the first nine digits are going to be 1 through 9 in some order, and we already know that

$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$

is a multiple of 9. So considering the first nine digits gives us no new information.

Step 4. The first digit has to be a multiple of 1. However, this isn’t helpful since every digit is a multiple of 1. So considering only the first digit gives us no new information.

I’ll continue this discussion with tomorrow’s post, where I’ll reduce the number of possible answers from 40,320 to only 576.

Pizza Hut Pi Day Challenge (Part 1)

I’m thinking of a ten-digit integer whose digits are all distinct. It happens that the number formed by the first n of them is divisible by n for each n from 1 to 10. What is my number?

To illustrate this idea, let me give an example that almost works: 9,632,581,470, a ten-digit number that uses each digit exactly once.

The first digit is 9, which is clearly a multiple of 1.
The first two digits form the number 96, which is clearly a multiple of 2: $96 = 2 \times 48$ .
The first three digits form the number 963, which is a multiple of 3: $963 = 3 \times 321$ .
The first four digits form the number 9632, which is a multiple of 4: $9632 = 4 \times 2408$ .
The first five digits form the number 96,325, which is a multiple of 5: $96,325 = 5 \times 19,265$ .
The first six digits form the number 963,258, which is a multiple of 6: $963,258 = 6 \times 160,523$ .
The first seven digits form the number 9,632,581, which is a multiple of 7: $9,632,581 = 7 \times 1,376,083$ .
Doesn’t work: $96,325,814 = 8 \times 12,040,726 + 6$ .
The first nine digits for the number 963,258,147, which is a multiple of 9: $963,258,147 = 9 \times 107,028,683$
The ten-digit number is a multiple of 10: $9,632,581,470 = 10 \times 963,258,147$ .

So, this number works for 9 of the 10 cases… but we need the number that works for all 10 cases.

In this series, I plan to discuss my solution of this problem using the rules of divisibility. I’ll start my solution with tomorrow’s post.

In the meantime, if you’d like to think about the solution on your own, I offer this green thought bubble to give you some time to think about it on your own.