Remainders

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Helenam

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We did some work on remainders as per the attached sheet. I seem to get the correct answers for all but the last although I’m not sure the second one is correct for the right reasons! Can anyone explain what I’m doing wrong?
Thanks. 19B5E3D1-44A2-4D7F-8647-2F5277BE0EDE.jpeg
 
Oops, just realised I forgot to add the 7 so answer is -4 which is still wrong! Cannot be a negative remainder.
 
I am not sure why you are doing the problems this way.

Here is what I would do for the 1st one.

a=5k+7, b=5

\(\displaystyle \dfrac {a}{b} = \dfrac {5k+7}{5} = \dfrac {5k+5+2}{5} = k+1 + \dfrac {2}{5}\), making r=2


Last one.
a=12k-7, b=3

The remainder for \(\displaystyle \dfrac {a}{b} = \dfrac{12k-7}{3} = the\ remainder\ for\ \dfrac{12k-7+9}{3} = \dfrac{12k+2}{3} = 4 +\dfrac{2}{3}\), making the remainder 2.

Note: Let's use 3 as an example. Let x be any integer. Then the remainder for x/3 will be the same for (x+3m)/3 for any integer m.
Division is repeated subtraction! If you have 29 marbles and keep taking away 3 marbles until you can't do that anymore, you'll have 2 marbles left over. If instead, you had 29 + 9 marbles you'll be able to take away 3 marbles three more times and still have 2 left over. Think about this.
 
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Steven G said:
I am not sure why you are doing the problems this way.

Here is what I would do for the 1st one.

a=5k+7, b=5

\(\displaystyle \dfrac {a}{b} = \dfrac {5k+7}{5} = \dfrac {5k+5+2}{5} = k+1 + \dfrac {2}{5}\), making r=2


Last one.
a=12k-7, b=3

\(\displaystyle \dfrac {a}{b} = \dfrac{12k-7}{3} = \dfrac{12k-7+9}{3} = \dfrac{12k+2}{3} = 4 +\dfrac{2}{3}\), making the remainder 2.

Note: Let's use 3 as an example. Let x be any integer. Then the remainder for x/3 will be the same for (x+3m)/3 for any integer m.
Division is repeated subtraction! If you have 29 marbles and keep taking away 3 marbles until you can't do that anymore, you'll have 2 marbles left over. If instead, you had 29 + 9 marbles you'll be able to take away 3 marbles three more times and still have 2 left over. Think about this.
Click to expand...
I follow the first example but the second one states that (12k-7)/3 = (12k-7+9)/3. How is this possible?
 
9/3 = 3. Adding any integer multiple of 3 (like 9) will not change the remainder when dividing by 3!

I did state this in the note.

Suppose you have ( 6k - 7)/3. This equals 2k -7/3. We will deal with the -7/3 in a minute.
Now (6k - 7 + 9)/3 = (6k+9-7)/3 = 2k+3 -7/9----regardless what the remainder is here, its the same as the line above!

However (6k+9-7)/3 = (6k + 2)/3 = 2k + 2/3. So the remainder is 2.

If you are dividing by 3 and just want to know the remainder, you can add or subtract any multiple of 3 that you like!

Division is repeated subtraction!
If you want to know the remainder when you divide 14 by 3, you can do the following.

14-3=11
11-3 = 8
8-3=5
5-3 =2
2 is the remainder!

If you want to know the remainder when you divide 14 by 3, you can add or subtract any multiple of 3 that you like to 14 and then do the following.

14+2*3 = 14 + 6 = 20
20-3 = 17
17- 3 = 14
14-3 = 11
11-3 = 8
8-3=5
5-3 =2
2 is the remainder!
Note how the red is exactly the same as above.
 
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Helenam said:
I am using the method our teacher showed us.
Click to expand...
Sorry
Steven G said:
9/3 = 3. Adding any integer multiple of 3 (like 9) will not change the remainder when dividing by 3!

I did state this in the note.

Suppose you have ( 6k - 7)/3. This equals 2k -7/3. We will deal with the -7/3 in a minute.
Now (6k - 7 + 9)/3 = (6k+9-7)/3 = 2k+3 -7/9----regardless what the remainder is here, its the same as the line above!

However (6k+9-7)/3 = (6k + 2)/3 = 2k + 2/3. So the remainder is 2.

If you are dividing by 3 and just want to know the remainder, you can add or subtract and multiple of 3 that you like!

Division is repeated subtraction!
If you want to know the remainder when you divide 14 by 3, you can do the following.

14-3=11
11-3 = 8
8-3=5
5-3 =2
2 is the remainder!

If you want to know the remainder when you divide 14 by 3, you can add any multiple of 3 that you like to 14 and then do the following.

14+2*3 = 14 + 6 = 20
20-3 = 17
17- 3 = 14
14-3 = 11
11-3 = 8
8-3=5
5-3 =2
2 is the remainder!
Note how the red is exactly the same as above.
Click to expand...
Got it. Apologies for my daftness. Your explanation is very clear and much appreciated. Thanks.
 
Helenam said:
I follow the first example but the second one states that (12k-7)/3 = (12k-7+9)/3. How is this possible?
Click to expand...
You are correct! However, I meant to say that the remainders for (12k-7)/3 and (12k-7+9)/3 are the same.
I edited my post. Good catch.
 
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