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Hello,
I am a college student and I'd like to say that this website is awesome! My teacher showed me this website earlier this week, since then I have been more interested in proofing mathematical problems.
Although this was a more visual proof, it was fun to discover patterns in powers of numbers and proof problems with these patterns.
I'd also like to say that English is not my first language, so please excuse my English :)
Furthermore, when solving these problem I noticed a pattern in all the powers table of single digit numbers: the last digit will always repeat on the 4th power. My question is: is there a mathematical proof for this (without using modular arithmetic)???
my email is: matheusgarcia28@gmail.com
I will be waiting for an answer! And once again, this is a very good website!
P.S: I am really sorry about the number of words used for this proof! After all, it was more about explaining the patterns.
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Problems
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In order to proof these problems, I used modular arithmetic. Modular maths is really simple to understand, for example, (10 MOD 9) is simple the remainder of 10/9 = 1. Modular arithmetic is explained in more depth in the article: http://nrich.maths.org/4350.
For which values of n will 2^n be a multiple of ten?
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64
2^7 = 128
2^8 = 256
2^9 = 512
2^10 = 1024
...
If N is a positive integer, then 2^n will never be a multiple of 10 because as we increase the power of 2, we are consequently doubling the previous value (starting with 1).
The only way to get a multiple of 10 by doubling a number is if we double some integer which has a 5 as its last digit, and quite clearly, we will never get such a number from the powers of 2 table.
Alternatives:
It's possible to generate multiples of 10 from powers of 2 if the power is LOGbase2(10M) where M is a positive integer.
Alternatively, from the table above we can spot a pattern: 2^n + 2^(n+2) will always be a multiple of 10.
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Why is 2^n + 3^n, where n is odd, always a multiple of 5?
3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = 243
3^6 = 729
3^7 = 2187
3^8 = 6561
3^9 = 19683
3^10 = 59049
If we compare the 2 powers table with the 3 powers table, we notice that there is a common pattern between them: the last digit -- on both powers table -- repeats after every 4th power, for example, the last digit of 2^5 and 2^9 would be the same.
From both power tables we can also see that the first odd power of 2 (2^1) = 4 and the first odd power of 3 (3^1) = 3, hence 2^1 + 3^1 = 5. Moreover, the second odd power of 2 (2^3) = 8 and the second odd power of 3 (3^3) = 27, thus 2^3 = 35.
Since the last digit will repeat every 4th power, the last digit of 2^1, 2^3, 3^1 and 3^3 will repeat throughout both powers tables. For this reason, we can use modular maths to deduce:
(2^n MOD 10) + (3^n MOD 10) = 5, where n is an odd integer.
What do we notice when n is a multiple of 4?
By using the explanation given above, we know that the last digit will repeat every fourth power. As (2^4 MOD 10) = 6, and (3^4 MOD 10) = 1, then the result of 2^n + 3^n, where n is a multiple of 4, will always have 7 as its last digit.
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For which values of n is 1^n + 2^n + 3^n even?
2^n will always be an even number,
3^n will always be an odd number,
1^1 will always be 1.
Therefore, 1 + (some odd number) + (some even number) will always be even. In conclusion, any positive integer value for n will make 1^n + 2^n + 3^n positive.
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For 1^n + 2^n + 3^n + 4^n, this will always be a multiple of 10 unless n is a multiple of 4.
We can proof this by using the assumption that the last digit of powers of 2,3 and 4 (or any single digit base) will repeat after every 4th power, and by using modular maths (yayyy!):
2^1 = 2, 3^1 = 3, 4^1 = 4 - adding these three results = 9
2^2 = 4, 3^2 = 9, 4^2 = 16 - adding these three results = 29
2^3 = 8, 3^3 = 27, 4^3 = 64 - adding these three results = 99
(notice that the last digits in powers of 4 repeat after every 2nd power)
Therefore:
(1 + 2^n + 3^n + 4^n) MOD 10 = 0, where n is a positive integer.
Now notice that, if n is a multiple of 4:
(1 + 2^4 + 3^4 + 4^4) MOD 10 = 4, hence this digit will be repeated for n=8, n=14 and so on...
We can apply the same explanation to 1^n + 2^n + 3^n + 4^n + 5^n, the result will always be a multiple of 5 unless n is a multiple of 4, in this case the last digit will always be 9.
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Surprising results:
2^n + 2^(n+2), where n is a positive integer
3^n + 3^(n+2), where n is a positive integer
---Notice the similarity between both powers table (3 and 2)---
4^n + 4^(n+1), where n is a positive integer
5^n - 3^n - 2^n, where n is odd (derived from 2^n + 3^n)
2^n + 8^n, where n is odd
Both results can be proofed by using the explanation (repetition of last digits) used previously.
[2^(3n + 3)] / 8^n, where n is a positive integer
3^(2n) / 9^n, where n is a positive integer
-- because 8 = 2^3, 4 = 2^2 9 = 3^2
I personally focused on multiples of ten, however, there are many more interesting results!