Divisibility problems

Today we went through some old contest problems involving divisibility. They looked similar, but not identical to these:

1. What number between 341 and 369 is divisible by both 3 and 5?
2. The six-digit number 54A319 is a multiple of 27. What digit is A?
3. Name three consecutive numbers, each less than 100, such that the smallest is divisible by 5, the next is divisible by 4, and the last is divisible by 3.
4. How many different counting numbers less than 161 are divisible by 4 or 5 or by both?
5. Suppose the number ABCDE is made from the digits 1,2,3,4,5. If the number CEA is divisible by 4, and the number EAB is divisible by 5, and the number ABD is divisible by 3, then what is ABCDE?

More modular arithmetic, and some logic puzzles

Modular arithmetic

Last week we came back to our problem from the previous two sessions about trying to hit all the points on a circle of n points. We made some progress on it, and I talked about the concept of relatively prime numbers (having no factors in common except 1), but I was sensing some restlessness, so this week we moved on to something different.

Logic puzzles

We started working on some simple logic problems, where you have, for example, three people with three different occupations, but you don't know which person has which occupation. Using the clues, you eliminate certain combinations until only one possibility remains.

The Centre for Innovation in Mathematics Teaching gives some examples of these. I used examples from this book.

We did several variations, where the correspondences were still small; just 3 or 4 things to match, but where the clues were more and more tricky.

For example, in one problem matching A, B, and C to X, Y, an Z, it was revealed that A was older than X. We talked about how this clue looks like it is giving only irrelevant information, but the inference you have to make is that if A is older than X, then A can't be X.

It was a lot of fun, and we will do some harder ones like this next week.

A problem from February's contest

There was an interesting problem on the last contest that required factoring to solve. In what I present here I have modified the problem slightly from the original, but if can solve this, you can solve the other.

We were given a rectangle divided into 4 smaller rectangles. The smaller rectangles all had integer sides, but they were not of equal size.

Three of the inner rectangles had areas given. The problem was to figure out the perimeter of the whole figure.

(The graphic is not necessarily to scale, so don't rely on that to solve it!)

One trick to beware of is that you may become focused on finding the area of the fourth rectangle. Although the process is essentially the same, don't forget at the end of your efforts to answer the question that was asked. In this case, we are looking for perimeter.

If you want to check your answer here, the perimeter you get should be 46.

More Modular Arithmetic

We spent our whole time exploring the points-on-a-circle problem from last time. The circle is just a graphical representation of modular arithmetic (a term I finally introduced after we got familiar with the mechanics). The problem is this:

If you are working in Integers Mod n, and you choose a number to count by, will you eventually hit every number in the set?

So, for example, if n = 6, you can count by 1 or 5, and eventually reach every number between 0 and 5, but 0, 2, 3, and 4 will all get stuck in cycles that don't cover all of the possible points.

Why is that?

As a general approach, I have decided to focus on developing intuition, and encouraging the students to make hypotheses. This worked really well today. They have come up with ideas, and were willing to share them even if they weren't sure they were correct. Then we all test together.

The first thing they noticed was that 1 and -1 (the last number before 0) always work. Then someone noticed that if n is even, no even numbers will ever hit any of the odd numbers. This is a key insight, which we talked through the correctness of.

We left off with the hypothesis that prime numbers always work, but we haven't proven it yet.

Eventually we will tie this back to a problem from December's contest, which ask the question: if you start at zero on a number line, and can only move to the right in steps of 3 or 10, what is the largest integer that you cannot get to?

Next week will be a contest.

Pascal's Triangle, Factoring, and Modular Arithmetic

Greetings,

I haven't updated in a while, in part because we've mostly been doing contests or going over contest problems.

Last week: Pascal's Triangle

Last week, in order to take a break from problems, we played around with Pascal's triangle. I based a lot of the lesson on Math is Fun's Pascal's Triangle page.

I printed some hex paper for us to use, and we built our own triangles.

First, we looked for patterns on the diagonal. We found the first two trivial ones. The third is a pattern we've looked at before: triangle numbers. The fourth we talked about briefly — tetrahedral numbers. It's like a 3-D version of triangle numbers. That's as far as we went with that.

We looked at some Sierpinski Triangle patterns, shading in evens for one iteration, and shading differently based on the remainder mod 3 for another. Here is my remainder mod 3 picture, where white is for 0, green is for 1, and blue is for 2.

Without using the term modulus, we talked about how for these triangles we don't need to know the value of the numbers in the hex to fill in the remainders, but this foreshadowed modular arithmetic, which I will return to below.

I briefly talked about how the Pascal numbers showed how many ways to flip coins heads or tails. For example, with a single coin flip, there is one way to get one head, and one way to get one tail, which maps to the second line in the triangle. With two flips, There is one way to get two heads, two ways to get one head and one tail, and one way to get two tails, which corresponds to the third line. We talked about how this means that the probability of one head and one tail is more probable than two of one kind, and so even though there are three possible outcomes, they aren't equally likely.

Relatedly, we talked about the number of ways to get to each hex if you start at the top and always go down.

Today: Factoring, and a little Modular Arithmetic

Prime factorisation

We started with finding prime factors in tree fashion:

We did 12, 51, and 156.

Then we worked on the following problems:

1. 12 has 6 factors. Find the smallest natural number with exactly 6 factors.
2. What is the largest odd factor of 90?
3. If N is the product of primes p, q, r, how many factors does it have?
4. How many numbers <= 100 have exactly 3 factors?

Factoring riddles

These do not necessarily have unique solutions. I got them from Bridges in Mathematics.

1. I am a common factor of 27 and 45. I am an odd number. When you multiply me by 3, you get a number greater than 10. What number am I?
2. I am a common factor of 36 and 48. I am also a factor of 30. I am an even number. I am divisible by 3. What number am I?
3. I am a common factor of 60 and 100. I am an even number greater than 4. I am divisible by 4. What number am I?
4. I am an odd number. I am a common factor of 135 and 210. I am greater than 7. What number am I?

Introducing Modular arithmetic

We recalled the work with Sierpinsky triangles and mod 3 arithmetic.

Then we worked on a couple of variations of the following problem:

Suppose we have a circle with 6 points on it labelled 0 through 5. If we start at the point labelled 0, which of the numbers between 1 and 5 can we keep adding and eventually land on all six points?

We also tried it for an 8 point circle, and started to make generalisations.

Next week we'll see what else what can find out about that problem.

Contest review plus sequencing games

Contest review

Today we went over the contest from November. It was tough! As usual, most people who took the contest nationally got 0, 1, or 2 of the 5 questions.

I won't list the problems here, because they are proprietary, but I'll categorise them.

The first was one of those rearrangement / calculation problems we talked about the first day. You can blast through it as it is, or you can rearrange it to make your calculations a little faster.

The second problem was a sequence puzzle where some clues are given and then you deduce the order of some objects. More on this below!

The third problem was to figure out a three digit number based on some clues about the digits, including divisibility.

The fourth involved calculating the area of a shape surrounding some other shapes.

The fifth was a cryptarithm. You can find plenty of examples of those at http://www.cryptarithms.com

We'll do more of all of those throughout the year.

Sequencing logic puzzles

This kind of puzzle is really fun. There are lots of these on the LSAT — the standardised test for getting into law school. To practise these, I got some related puzzled from the Mindware game LogicLinks. They have samples from the four leveled books:

• Level A
• Level B
• Level C
• Level D

We worked through these until we ran out of time!

Divisibility by 3 and a Contest

Sums and differences again: remainders

Last time we talked about using sums and differences to tell if a number is divisible by 7.

For example, 781 is not divisible by 7: $781=770+11$, and we know that 770 is divisible by 7 and that 11 is not.

A similar kind of logic works with remainders.

• What is the remainder (R) when you divide 465 by 7?

  $465=420+45$.

  $45÷7$ → R3.

• What is the remainder if you divide (13 + 283) by 7?

• Method 1: $13+283=296=280+16$. $16÷7$ → R2
• Method 2: $13÷7$ → R6; $283÷7$ → R3; $6+3=9$; $9÷7$ → R2, so $(13+283)÷7$ → R2

The divisibility by 3 "trick":

If the digits in a number sum to a multiple of 3, then the number is divisible by 3.

How do we know this works?

First, see if there are any obvious exceptions: does it work for the first few multiples of 3 you can think of? How about the first few non-multiples of 3?

Okay, so you may have persuaded yourself that there is something to it, but does this prove it always works?

Of course not!

Here's the reason it works:

$10÷3$ → R1

So if you have, for example 54, $54=50+4$. You know that 50 will have a remainder of 1 when divided by 3, and that will happen 5 times. As far as remainders are concerned, this is just like adding 5.

We had to stop here to start...

Our first contest

which was really fun!

We'll go over the solutions, not next week, but the week after.