Question 1151775
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I'll do part (a) to get you started.


p is some prime number such that p > 3.


We are dividing this prime by 6 to get
p/6 = q+r/6
q = quotient
r = remainder
the quotient and remainder are integers


Multiply both sides by 6
p = 6q + r
The remainder r can be any one of these values {0,1,2,3,4,5}
If r is even, r = 0,2,4, then we can show that p itself is even
For instance, if r = 2, then
p = 6q+r
p = 6q+2
p = 2(3q+1)
p = some multiple of 2
but p has to be odd in order for it to be a prime number larger than 3.
So we can rule out any cases when r is even.


This leaves r = 1,3 or 5.
But if r = 3, then,
p = 6q+r
p = 6q+3
p = 3(2q+1)
showing that 3 is a factor of p, therefore invalidating p being a prime number
A prime number only has factors of 1 and itself
We can rule out r = 3.


So r = 1 or r = 5 are the only remainders possible.
Here is a table of some primes mod 6
<table border = "1" cellpadding = "5">
<tr><td>p</td><td>p mod 6</td></tr>
<tr><td>5</td><td>5</td></tr>
<tr><td>7</td><td>1</td></tr>
<tr><td>11</td><td>5</td></tr>
<tr><td>13</td><td>1</td></tr>
<tr><td>17</td><td>5</td></tr>
<tr><td>23</td><td>5</td></tr>
<tr><td>29</td><td>5</td></tr>
<tr><td>31</td><td>1</td></tr>
<tr><td>37</td><td>1</td></tr>
<tr><td>41</td><td>5</td></tr>
<tr><td>43</td><td>1</td></tr>
</table>

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