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September 23rd, 2018, 03:17 AM   #1
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Unhappy 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.
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September 23rd, 2018, 05:25 AM   #2
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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.
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September 23rd, 2018, 09:44 AM   #3
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Quote:
Originally Posted by skipjack View Post
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.
Well, that is what I thought,
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 .
DecoratorFawn82 is offline  
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