Power mad!
Powers of numbers behave in surprising ways...
Take a look at the following and try to explain what's going on.
Work out $2^1, 2^2, 2^3, 2^4, 2^5, 2^6$...
For which values of $n$ will $2^n$ be a multiple of $10?$
For which values of $n$ is $1^n + 2^n + 3^n$ even?
Work out $1^n + 2^n + 3^n + 4^n$ for some different values of $n$.
What do you notice?
What about $1^n + 2^n + 3^n + 4^n + 5^n?$
What other surprising results can you find?
Here are some suggestions to start you off:
$4^n + 5^n + 6^n$
$2^n+3^n$ for odd values of $n$
$3^n + 8^n$
$2^n + 4^n + 6^n$
$3^n + 5^n + 7^n$
$3^n - 2^n$
$7^n + 5^n - 3^n$
Can you justify your findings?
You may also like to take a look at Big Powers.
Click here for a poster of this problem.
What is the impact of the units digits on the product of two or more numbers?
e.g. what do you notice about the units digits in all the answers to the following:
What can you say about the patterns of the units digits of powers of 2, 3, 4, 5...?
Here is a really nice solution submitted by Oliver from Loreto College:
Assuming n is an integer, there are no values of n so that $2^n$ is a multiple of 10 because $2^n$ doesn't contain any necessary factors of 5.
The unit digits of $2^n$ for n=1,2,3... are 2, 4, 8, 6 then repeat. For $3^n$ they go 3, 9, 7, 1 then repeat. If n is odd, the units that are being added are either 2 + 3 or 7 + 8, which both end in 5. So $2^n + 3^n$ where n is odd always ends in 5. This is a stronger conclusion than saying 'it's a multiple of 5' as a multiple of 5 can also end in 0.
If n is a multiple of 4 the units being added are 6 + 1 so in this case it will always end in 7.
$1^n + 2^n + 3^n$ is even for all values of n. This is obvious because $1^n$ and $3^n$ are always odd and $2^n$ is always even, and odd + odd + even =even.
$1^n + 2^n + 3^n + 4^n$ is a multiple of 10 for when 4 does not divide n:
The unit digits of the powers of 1, 2, 3 and 4 are:
| x^1 | x^2 | x^3 | x^4 |
| 1 | 1 | 1 | 1 |
| 2 | 4 | 8 | 6 |
| 3 | 9 | 7 | 1 |
| 4 | 6 | 4 | 6 |
Summing down the columns gives 10, 20, 20 and 14. This shows that when n is not divisible by 4, the last digit of $1^n + 2^n + 3^n + 4^n$ is 0 so it's a multiple of 10. (When n is divisible by 4, the last digit is 4).
$1^n + 2^n + 3^n + 4^n + 5^n$ ends in 5 for n not divisible by 4. This is obvious if we consider the previous result because $5^n$ ends in 5 for all n, so adding this to the multiples of 10 will give a final digit of 5.
Ryan from Renaissance College Hong Kong had another good explanation for why $2^n$ cannot be a multiple of 10:
| Power of 2 | Answer | Units Digit |
| 1 | 2 | 2 |
| 2 | 4 | 4 |
| 3 | 8 | 8 |
| 4 | 16 | 6 |
| 5 | 32 | 2 |
| 6 | 64 | 4 |
| 7 | 128 | 8 |
| 8 | 256 | 6 |
For which values of n will $2^n$ be a multiple of 10?
As can be seen from the following table, the unit digits of the powers of
two are in a repetitious pattern of 2, 4, 8, 6, 2, 4, 8, 6”¦
All multiples of 10 have a unit digit of 0. However, as seen from the
pattern, none of the powers of 2 have their unit digit ending in a 0.
Therefore, no power of 2 is a multiple of 10.
Alexander from University College School noticed the following about the extension questions:
If the power n in $4^n + 5^n + 6^n$ is an odd number, then the last digit of this sum is 5; if the power n is even then the last digit of the sum is 7.
$3^n + 8^n$: For consecutive values of n (n = 1, 2, 3, ”¦) the last digit of the sum goes in a pattern of 1, 3, 9, 7.
$2^n + 4^n + 6^n$: For consecutive values of n (n = 1, 2, 3, ”¦) the last digit of the sum goes in a pattern of 2, 6, 8, 8.
$3^n +5^n +7^n$: For consecutive values of n (n = 1, 2, 3, ”¦) the last digit of the sum goes in a pattern of 5, 3, 5, 7.
$3^n-2^n$: For consecutive values of n (n = 1, 2, 3, ”¦) the last digit of the sum goes in a pattern of 1, 5, 9, 5.
Can you explain why this might be true?
Ryan also had some interesting ideas about how he could extend his results:
The unit digit of $1^n$ is 1, 1, 1, 1”¦
The unit digit of $2^n$ is 2, 4, 8, 6, 2, 4, 8, 6,...
The unit digit of $3^n$ is 3, 9, 7, 1, 4, 9, 7, 1,...
The unit digit of $4^n$ is 4, 6, 4, 6”¦
The unit digit of $5^n$ is 5, 5, 5, 5”¦
The unit digit of $6^n$ is 6, 6, 6, 6”¦
The unit digit of $7^n$ is 7, 9, 3, 1, 7, 9, 3, 1,...
The unit digit of $8^n$ is 8, 4, 2, 6, 8, 4, 2, 6,...
The unit digit of $9^n$ is 9, 1, 9, 1”¦
The unit digit of $10^n$ is 0, 0, 0, 0”¦
Thus the unit digits of the powers of any number repeat in a cycle four digits long, as all numbers end with one of the above 10 digits.Let the pattern of the unit digits of the powers of x be a, b, c, d... and let The pattern of the unit digits of the powers of another number y be e, f, g, h”¦
As a result, the sum/difference of the unit digits of the powers of x and y are in a repeating pattern of: a + e, b + f, c + g, and d + h, or a - e, b - f, c - g and d - h.
As a result, no matter what combination of powers, adding or subtracting, the unit digits are always in a pattern of 4 repeating numbers.
Well done everyone!
Why do this problem?
This problem offers practice in working with indices to develop fluency, while providing an intriguing context to discover patterns and find justifications.
Possible approach
This printable worksheet may be useful: Power Mad
"Work out and write down the powers of $2$ from $2^1$ up to $2^8$." Give students a short time to do this, perhaps using mini-whiteboards.
"What do you think would be the last digit of $2^{100}$?" Give students time to discuss this with their partner before sharing ideas and justifications.
"Are there any powers of two that are multiples of $10$?" "No, because a power of 2 has to end in a 2, 4, 6 or 8, and a multiple of 10 ends in a 0".
For the next part of the lesson, you could divide the class into pairs or small groups, and give each group one of the following to work on:
- Work out $2^n+3^n$ for some different odd values of $n$.
What do you notice?
What do you notice when $n$ is a multiple of $4?$
- For which values of $n$ is $1^n + 2^n + 3^n$ even?
- Work out $1^n + 2^n + 3^n + 4^n$ for some different values of $n$.
What do you notice?
- Work out $1^n + 2^n + 3^n + 4^n + 5^n$ for some different values of $n$. What do you notice?
When students have finished working on their question and justified their findings, invite them to look for similar results of their own. Here are a few suggestions that they could explore for different values of $n$:
$3^n + 8^n$
$2^n + 4^n + 6^n$
$3^n + 5^n + 7^n$
$3^n - 2^n$
$7^n + 5^n - 3^n$
To finish off, students could present their findings to the rest of the class, with emphasis on clear explanations to justify that the patterns they have found will continue for all values of $n$.
Key questions
What patterns can you find in the units digit of ascending powers of 2, 3, 4...?
Possible support
You might suggest that students draw up 'power tables' so that the cyclical nature of the units digits becomes apparent.
Possible extension
This is an open ended activity which already offers plenty of opportunities for extension work.
The Stage 5 problem Tens takes the ideas in this problem and treats them in a more formal way, encouraging the use of Modular Arithmetic notation.