If you change the prime number 13 to base 7 you get 16. How can 16 be a prime number? Is it that 16 base 7 is a prime number in base 7 math? Can you give an example of how this could be true?
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A prime number has no factors besides itself and 1. What factors would 16 in base-7 suddenly have that 13 in base-10 doesn’t have?
Changing the base doesn’t change the number, just how it’s expressed. 13 is 13 no mater how you write it out.
Changing the base doesn’t change the number, just how it’s expressed. 13 is 13 no mater how you write it out.
A prime number has no factors besides itself and 1. What factors would 16 in base-7 suddenly have that 13 in base-10 doesn’t have?
Count out this many objects:
O O O O O O O
O O O O O O
In base ten, it’s written 13.
In base seven, it’s written 16.
Either way, that number of objects is prime.
No matter the base, you cannot take this many objects and come up with multiple factors for it.
Count out this many objects:
O O O O O O O
O O O O O O
In base ten, it’s written 13.
In base seven, it’s written 16.
Either way, that number of objects is prime.
No matter the base, you cannot take this many objects and come up with multiple factors for it.
Basically, changing a number’s base doesn’t change anything about a number’s properties. The only thing that changes is how we write it.
>How can 16 be a prime number?
Because 16 is not even in base 7. The reason that we only have to check the last digit in base 10 to determine whether a number is even or not is because 10 itself is also even and therefore adding any number of 10s or its powers will not affect a number’s evenness – this is not the case in base 7.
10 looks even to us because we’re used to base 10, but if it is in base 7 and we subtract 2 from it – we get 5, which is odd, and looks odd to us in both base 7 and base 10.
Note that I’ve largely talked about evenness rather than primeness, and this is because I assume that your question about 16 in base 7 comes from it “looking” even (and therefore not prime) because we’re used to base 10.
Basically, changing a number’s base doesn’t change anything about a number’s properties. The only thing that changes is how we write it.
>How can 16 be a prime number?
Because 16 is not even in base 7. The reason that we only have to check the last digit in base 10 to determine whether a number is even or not is because 10 itself is also even and therefore adding any number of 10s or its powers will not affect a number’s evenness – this is not the case in base 7.
10 looks even to us because we’re used to base 10, but if it is in base 7 and we subtract 2 from it – we get 5, which is odd, and looks odd to us in both base 7 and base 10.
Note that I’ve largely talked about evenness rather than primeness, and this is because I assume that your question about 16 in base 7 comes from it “looking” even (and therefore not prime) because we’re used to base 10.
Can you find a factor of 16-base-7 in base-7?
For possible numbers, we have 1, 2, 3, 4, 5, 6, 10, 11, 12, 13, 14, 15, 16. Obviously 1 and 16 are out, since those are the prime factors.
What is 16 / 2 in base 7? Since pretty much all calculators operate in base-10 by default, we have to do this manually:
16 – 2 is 14.
14 – 2 is 12.
12 – 2 is 10.
10 – 2 is 5.
5 – 2 is 3.
3 – 2 is 1.
1 – 2 is (-1), so 2 cannot be factor of 16.
Since 2 isn’t a factor, neither can 4, 6, or 11, 13, or 15, since 2 is a factor of those numbers (do you know why?).
So we have 3, 5, and 10, 12, and 14 left as possible factors.
16 – 3 is 13.
13 – 3 is 10.
10 – 3 is 4.
4 – 3 is 1. 3 cannot be a factor.
16 – 5 is 11.
11 – 5 is 3. 5 cannot be a factor.
16 – 10 is 6. 10 cannot be a factor.
16 – 12 is 4. 12 cannot be a factor.
16 – 14 is 2. 14 cannot be a factor.
There are no factors of 16-base-7. Ergo, 16-base-7 is a prime number, just as 13-base-10 is.
Let’s look at a non-prime example; 13-base-7. It looks like a prime, but it isn’t:
Let’s try 2 as a factor.
13 – 2 is 11.
11 – 2 is 6.
6 – 2 is 4.
4 – 2 is 2.
2 – 2 is 0. So 2 is a factor of 13-base-7, as well as 5 (the number of times 2 factors into 13-base-7). Of course this works because 13-base-7 is 10-base-10, so 2 times 5 makes 10.
Can you find a factor of 16-base-7 in base-7?
For possible numbers, we have 1, 2, 3, 4, 5, 6, 10, 11, 12, 13, 14, 15, 16. Obviously 1 and 16 are out, since those are the prime factors.
What is 16 / 2 in base 7? Since pretty much all calculators operate in base-10 by default, we have to do this manually:
16 – 2 is 14.
14 – 2 is 12.
12 – 2 is 10.
10 – 2 is 5.
5 – 2 is 3.
3 – 2 is 1.
1 – 2 is (-1), so 2 cannot be factor of 16.
Since 2 isn’t a factor, neither can 4, 6, or 11, 13, or 15, since 2 is a factor of those numbers (do you know why?).
So we have 3, 5, and 10, 12, and 14 left as possible factors.
16 – 3 is 13.
13 – 3 is 10.
10 – 3 is 4.
4 – 3 is 1. 3 cannot be a factor.
16 – 5 is 11.
11 – 5 is 3. 5 cannot be a factor.
16 – 10 is 6. 10 cannot be a factor.
16 – 12 is 4. 12 cannot be a factor.
16 – 14 is 2. 14 cannot be a factor.
There are no factors of 16-base-7. Ergo, 16-base-7 is a prime number, just as 13-base-10 is.
Let’s look at a non-prime example; 13-base-7. It looks like a prime, but it isn’t:
Let’s try 2 as a factor.
13 – 2 is 11.
11 – 2 is 6.
6 – 2 is 4.
4 – 2 is 2.
2 – 2 is 0. So 2 is a factor of 13-base-7, as well as 5 (the number of times 2 factors into 13-base-7). Of course this works because 13-base-7 is 10-base-10, so 2 times 5 makes 10.
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