Intriguing Indices
Summary
The aim of this lesson is to review, explain and practice the rules of indices. The exercises provided are designed so as to offer a natural opportunities to discuss general rules. Thus the format of this session should ideally alternate between individual/group problem solving and “class discussion”, meaning those times when the students’ attention is focused on the session leader, who along with demonstrating rules/techniques, engages the class by asking related questions. The class discussion areas of the lesson plan are highlighted in purple, and questions in red.
Students may require support with starting some of the exercises. It’s good to be ready with hints and to start a class discussion at any time. In particular, it is possible that this lesson occurs before the students study indices in school. If this is the case, the session leaders could provide a few extra simple numerical examples for each rule before going into the more abstract exercises proposed here. Still, the purpose is not to drill the rules of indices as much as to emphasize the common sense ideas on which they are based.
Resources
- Calculators for may be useful.
- Activity Sheet per student
- Solution Sheets for tutors
Questions/Suggestions?
If you plan to use this material, or if you would write to send us feedback, please email
Or write to:
Kieran Cooney, Anca Mustata
School of Mathematical Sciences, School of Mathematical Sciences,
UCC, Cork UCC, Cork
Intriguing Indices
1.Double Trouble
2. Multiplication Rule
a) Calculate:
b) Find the smallest positive integer such that the number is a perfect power
(that is, the power of some integer).
c) Which is larger: or ?
3.Division Rule
a) Fill in the following table:
-4 / -3 / -2 / -1 / 0 / 1 / 2 / 3 / 4b) Simplify:
4. A guessing game.
a) If you play well, you should be able to guess your number in less than moves.
Fill in the smallest number that you think will always work. Explain your choice.
b) As 2nd player, I’ll use my minimal-guess strategy demonstrated in class. What mystery numbers would you choose to stall me as much as possible?
c) At most how many trials would you need to guess a number between 1 and 1000?
5. Guessing game 2D
Two chess players on a break play a guessing game.
The 1st player places an imaginary rook on one of the squares.
The 2nd player touches any point on the board. If the point is inside the same square as the imaginary rook, or on its border, the game is over. Otherwise, the 1st player indicates the relative position of the rook by UP/DOWN and LEFT/RIGHT.
a) What’s the largest possible number of “moves” in this game?
b) The same question assuming the board was made out of 4096 small black-and-white squares.
6. Multiple Indices:
a) Is this number a perfect square:
Find such that: .
7. Trick-or-Treat
You are offered a choice of one of two boxes. Each box carries a label stating the number (or fraction!) of identical goodies inside. Compare the labels and pick one:
8. Sums of exponents
a) Solve for
b) Show that is always divisible by 13.
c) Show that is always divisible by 7.
9. Lagrange's four-square theorem
Intriguing Indices – Practice Questions
1.Simplify:
2. Solve for :
3. Solve for when ():
4. Find the largest integer such that .
5. Compare each of these pairs of numbers
6. a) Write as the sum of 2 squares.
b) Write as the sum of 2 squares.
c) Write as the sum of 2 cubes.
d) Write as a sum of two square numbers.
e) Write as a sum of two cube numbers.
7. a) Calculate
b) Write as a sum of five perfect squares.
8. How many digits does the number have?
9. Digit Tricks
a) Find the last digit of
b) Find the last digit of
c) Find the last digit of
d) Find the last digit of
e) The number ends in exactly how many zeroes?
Intriguing Indices – Solutions and Class Discussion
1.Double Trouble
Solution and class discussion: It is not hardto conclude that there are 64 leaves, but it is interesting to discuss the different ways to reach this number:
i) The tree can be split in 2 equal parts by left-right symmetry. The leaves in each part are roughly arranged in 4 groups of 8 leaves each:
ii) Starting from the tree trunk, count how many branching points there are en route to a leaf. Each branching in 2 doubles the number of leaves. There are 6 branching points so
leaves in total. One can also place a number at each branching point forming this sequence:
1,2,4,8,16,32,64.
Here 6 is an index or an exponent. If we are calling it an index, we must be careful as there are other things in maths referred to as indices as well. 64 is called a power of 2 because we can write it as a product of 2s only.
All the numbers above are powers of 2: indeed The first expression above becomes:
Do you notice anything interesting about the indices?They sum as
And no wonder:
In both cases, 2 is multiplied by itself 6 times.
Multiplication Rule
2. Multiplication Rule
a) Calculate:
b) Find the smallest positive integer n such that the number is a perfect power
(that is, the power of some integer).
c) Which is larger: or ?
Solution: a)
b) Hint: write each of the numbers 48 and 54 as products of powers and then combine: note that the order of the factors in the product doesn’t matter (commutativity of multiplication).
is missing a factor of 3 to become = a perfect power.
The smallest number we can find is
Can you explain why ?
c) .
So the second number is bigger.
3. Division Rule
a) Fill in the following table:
-4 / -3 / -2 / -1 / 0 / 1 / 2 / 3 / 4b) Simplify:
Solution: Let’s discuss the pattern:
/ / / / / / / / / …1 / 2 / 3 / 4 / 5 / 6 / 7 / 8 / 9 / …
Above, we have drawn a table where we keep multiplying by itself and keep count of how many times we have on the bottom. If we take any term on the top line and multiply it by , we notice that we move one place to the right, which is the same as adding one on to the index. We could extend this to show the multiplication rule! What if we divide a term by Well there will now be one less in the product, so it will shift to the by one place. We subtract one from the index. What if we divide by ? We will move two places to the left, subtracting two from the index. If we divide by , we will move places to the left, subtracting from the index. So
Division Rule
Hence. We can look at this in different ways. Starting at , let’s multiply it by and see where it takes us on the chart. . So multiplying by doesn’t change the number at all. But there is only one number that does this, 1! Hence .
Another way to explain this is to apply the formula in the case when
So And this is true for all … or is it?
From above, we would say that . But , and it’s fairly easy to see that . But thenshouldn’t ? So what’s wrong? This is an inconsistency in arithmetic: We say that is undefined, there is no number that represents it because it doesn’t obey all rules the way we would like it to. This happens sometimes in mathematics, for example what is ? Note that this is exactly why doesn’t work for 0.
Negative exponents
Assuming that we can fill in the table as below: To get to negative exponent, we need to move to the left of the chart, and this involves division. So if we start at a number like , and keep moving left by dividing we see
... / / / / 1 / / / / …… / -3 / -2 / -1 / 0 / 1 / 2 / 3 / …
And we can keep moving back like this. Thus
Another way to see this is to take in the formula .
1,875
4. A guessing game.
Play this in pairs. The 1st player thinks of a number between 1 and 63. The 2nd player tries to guess it. After every guess, the 1st player has to state (truthfully!) whether his/her chosen number is bigger or smaller than the one just heard. This goes on until the mystery number is found.
The players take turns being first and each time, the number of guesses is tallied.
The player who needed fewer guesses wins.
a) If you play well, you should be able to guess your number in less than moves.
Fill in the smallest number that you think will always work. Try it out!
b) As 2nd player, I’ll use my minimal-guess strategy demonstrated in class. What strategy would you choose to stall me as much as possible?
c) At most how many tries would you need to guess a number between 1 and 1000?
Solution: After the pairs have played, have a class session with a volunteer student as 1st player and the lecturer/tutors as 2nd player. The volunteer can write the mystery number on a piece of paper and circulate it to the class.
Here is the 2nd player’s strategy:
Step 1: Guess 32.
Step 2: Add or subtract 16 to your previous guess: “bigger”= and “smaller” =
Step 3: Add or subtract 8 to your previous guess.
Step 4: Add or subtract 4 to your previous guess.
Step 5: Add or subtract 2 to your previous guess.
Step 6: Add or subtract 1 to your previous guess.
Less than 7 moves (at most 6 moves) are needed because
Also note that
So if we used this strategy, we’d need 7 moves to get to 64.
b) Here is the 2nd player’s cheat sheet for the first 5 steps:
32
16 48
8 24 40 56
4 12 20 28 36 44 52 60
2 6 10 14 18 22 26 30 34 38 42 46 50 54 58 62
It’s pretty clear that all the odd numbers will fall in the last step. How can we prove this? We’ll return to this after discussing binary basis.
c) so 10 guesses should suffice. One may introduce the logarithm notation: is equivalent to
In general, answers the question: raised to which power equals ?
The number of trials we’ll need to guess a number between 1 and is at most
5. Guessing game 2D
Two chess players on a break play a guessing game.
The 1st player places an imaginary rook on one of the squares.
The 2ndplayer touches any point on the board. If the point is inside the same square as the imaginary rook, or on its border, the game is over. Otherwise, the 1st player indicates the relative position of the rook by UP/DOWN and LEFT/RIGHT.
a) What’s the largest possible number of “moves” in this game?
b) The same question assuming the board was made out of 4096 small black-and-white squares.
Solution: a) This is a very short game indeed, with only 3 moves needed. We first divide the board into 4 equal squares and touch the point in the centre, then move to the square indicated by the UP/DOWN and LEFT/RIGHT hints, divide this into 4 equal squares, move on and again divide by 4 til at the last step we get squares.
Looking at the total number of small square this is explained by Looking at side lengths: Perhaps this is a good moment to note that
So is both a perfect square and a perfect cube.
Multiple Indices Rule:
Indeed, from our definition of exponents
=
times times times times
times
b) so we will need 6 moves to finish the game.
6. Multiple Indices:
a) Is this number a perfect square:
Find such that: .
Hints: Write everything in terms of powers of 3 and 2.
Solution:
c) By playing with the exponents, we get the equation so
A negative fraction. Ugh.
What is ? Can you explain why? Does always exist?
What if we get rid of the annoying denominator?
So if we square , we will get , so that means that
Note that this entire argument goes awry if Indeed, then how could a negative number equal a positive ? So is not defined when
What about ? If then by raising it to the -th power we see that
7. Treat-or-Trick
You are offered a choice of one of two boxes. Each box carries a label stating the number (or fraction!) of identical goodies inside. Compare the labels and pick one:
a) or ? b) or ?
c)or ? d) or ?
e) or ? f) or ?
a) The emphasis here is for the students to realise that if two numbers greater than 1 have the same bases but different indices, we can compare the results by comparing the indices:
To simplify , we start from the top and work down. so
We can do this for bigger and bigger “telescopes” of exponents, as long as we start from the top and work down. On the other hand, so those who like Halloween stuff would choose the 1st box as
The general rule:
If the basis then the number with the larger index is the larger
forwhenever
because
b) This is weird: you’re getting fractions of the goodies:
And so the second number is larger. This can also be seen in the following way:
Because half of a positive quantity is less than the whole.
So for numbers between 0 and 1, the situation is reversed:
forwhenever
because
c)This reduces to If two numbers have the same exponents but different bases (the thing we multiply by itself), then will the number with the bigger base always be bigger?
d) This reduces toversus. Even though is the same on both sides and we have because 1 fifth of something is more than 1 sixth.
To recap, one always has to be careful when negative signs are parts of inequalities...
There’s also the question as to the actual content of the boxes: are these fractions of a mountain of candy? Then they may be worth something...
e) The idea for both questions is to make either the index or base the same, and then compare using ideas we have built up already. Notice that
f) Observing that divides each of the indices, we write the 1st number as and the 2nd as As , the 2nd number is bigger.
8. Lagrange's four-square theorem
Solution: a) Calculate.
b) Hints: Difference of two squares and factor out a square whenever possible:
c) Hints: You can factor out a square: .
d) Same trick: and use b).
e) Same trick: or
and
9. Sums of exponents
a) Solve for
b) Show that is always divisible by 13.
c) Show that is always divisible by 7.
Solution: a) Notice that is a factor of each of the terms on the left hand side. So we use distributivity and take it outside.
Hence and
b) First, we make sure that the base is the same, so our expression becomes . Now we can factor out to get . As it is a multiple of 13, is divisible by 13.
c) The problem here is that we can’t simplify everything, as we are working with bases of 2 and 5. So we deal with each part separately. In the end, we should have something like
So the expression is always divisible by 7.
Supplementary Exercises
1.Simplify:
Answer: a) b) c)
2. Solve for :
Answer:
3. Solve for when ():
Answer:
4. Find the largest integer such that .
Solution:
, so . The largest such is .
5. Compare each of these pairs of numbers
Solution: a) . So the second term is larger.
b) Upon simplification, the two terms are . and . As , the first term is larger.
c) b) c)
6. a) Write as the sum of 2 squares.
b) Write as the sum of 2 squares.
c) Write as the sum of 2 cubes.
d) Write as a sum of two square numbers.
e) Write as a sum of two cube numbers.
Solutions: a)
b)
c)
d) Let’s do exactly the same thing here as we did for the last one. , so
e) There is a slight difference here because we are using cubes, but let’s use the same methodology anyway. , so
7. a) Calculate
b) Write as a sum of five perfect squares.
Solution:
a) The answer should be 2011
b)
There is a bit of clever manipulation here that the students might struggle with. If they remember everything that has been done so far, (and use it!) they might get it.
8. How many digits does the number have?
Solution:. So there are digits.
9. Digit Tricks
a) Find the last digit of
b) Find the last digit of
c) Find the last digit of
d) Find the last digit of
e) The number ends in exactly how many zeroes?
Solution: a) This is obviously 5.
b) There is no way we can do this by actually figuring out what is; there will be too many digits. But we can be clever and only use 1 digit. If we go through the first few values of looking only at the last digit, we get a pattern: We know that as soon as we get to 1 in our pattern that the pattern has ended. This is because the next number will be our starting number. This pattern repeats after 4 terms. So the last digit depends on what remainder 2012 has upon division by 4. Since 4 divides 2012, it’s remainder is zero and ’s last digit is 1.
Note: Some patterns don’t end in 1, just look at the next example.
c) The pattern goes like so has a length 2. So we need to find out wether is even or not. As it is even, then has a last digit 1.
d) The pattern for the last digit of goes like so the pattern has length 4. So we need to find the remainder of upon division by 4. is divisible by 4, so the remainder of upon division of 4 is 2. Hence, the last digit of is the second term in the pattern, 4.
e) Write each of the factors of 1000! as a product of prime numbers. It suffices to find how many times 5 occurs in that product. Indeed, 10=2\times 5 and 2 will occur more times than 5 in 1000! (because of all those even numbers!), so each time a 5 occurs, it can be paired with a 2 to form a 10.
Among the numbers in the list 1, 2, ..., 1000 there are 200 multiples of 5, there are 40 multiples of 25, there are 8 multiples of 125 and 1 multiple of 625. All in all, 5 will occur 200+40+8+1=249 times.