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September 23rd, 2018, 02:17 AM  #1 
Member Joined: Sep 2014 From: Sweden Posts: 94 Thanks: 0  If f gives remainder 3 by division with 6, what remainder do 2^f^2 + 5 give?
If f gives remainder 3 by division with 6, what remainder do we get when we divide 2f^2 + 5 by 6? So I think I have made an attempt to solve this at least, but are not really sure if it is correct. $\displaystyle f = bq + r$ where $\displaystyle q \in \mathbb{Z}, 0 \le r< b, \ r \in \mathbb{Z}$ We have that $\displaystyle b = 6$ and $\displaystyle r = 3$. Goal: $\displaystyle 2f^2 + 5 = 6q_{1} + 5$. The goal is to write it in this form to know what remainder it gets when dividing by 6. Put in $\displaystyle f = 6q +3$ into $\displaystyle 2f^2 + 5$. Calculate LHS: $\displaystyle 2f^2 + 5 =$ $\displaystyle 2(6q+3)^{2} + 5 =$ $\displaystyle 2(6^{2}q^{2} + 2 * 3 * 6 * q + 3^{2}) + 5 =$ $\displaystyle 2 * 6^{2}q^{2} + 2 * 2 * 3 * 6 * q + 2 * 3^{2} +5 =$ $\displaystyle 6(2 * 6 * q^{2} + 2 * 2 * 3 * q + 3) + 5$ So the number within the parenthesis must be an integer. Now we have the expression on the form $\displaystyle f = 6q_{1} + r$ that was the goal. So $\displaystyle r = 5$ So this is my conclusion but to me it seems wrong because that is what is given from the beginning in the assignment. So I am not really sure this is correct. 
September 23rd, 2018, 04:25 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,969 Thanks: 2219 
Your work is correct, but can be shortened considerably. If $f = 6q + 3$, 3 divides $f$, and so 6 divides $2f^2$. Hence the remainder on dividing $2f^2 + 5$ by 6 is 5. 
September 23rd, 2018, 08:44 AM  #3  
Member Joined: Sep 2014 From: Sweden Posts: 94 Thanks: 0  Quote:
but my school probably wants me to stick with this method (for now at least) for me to show that I have understood how it works. Thanks for clarifying .  

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