What's the easiest and fastest way to do this classic math prob?

<p>When the positive integer k is divided by 7, the remainder is 6. What is the remainder when k+2 is divided by 7?</p>

<p>a) 0
b) 1
c) 2
d)3
e)4</p>

<p>Now I got the answer right, but I did it by guessing and checking a few numbers until I found that k could equal 15.
I went to collegeboardwebsite and there explanation is:</p>

<p>Since the positive integer k leaves a remainder of 6 when divided by 7 it follows that k can be written as k=7q+6 where q is some nonnegative integer. Thus k+2 can be written as k+2=7q+8 or 7(q+1)+1. Since q+1 is a positive integer and k+2= 7(q+1)+1 it follows that k+2 leaves a remainder of 1 when divided by 7.</p>

<p>what is ur own way ?? or which is better?</p>

<p>Idk about all that yucky algebra, but all I did was suppose k was 13…</p>

<p>dividing is the same as subtracting repeatedly…remainder is 6 so you cant subtract 7. remainder becomes 8 when you add 2 so you can subtract 7 and get a 1 remainder</p>

<p>7 divided by 7 is 1…0 remainder 7<em>1+0=7
13 divided by 7 is 1…6 remainder, 7</em>1+6=13
14 divided by 7 is 2…0 remainder 7<em>2+0=14
15 divided by 7 is 2…1 remainder 7</em>2+1=15</p>

<p>you shouldve understood the concept behind this question when you first learned division…
1 2 3 4 5 6 7
1 2 3 4 5 6 7
1 2 3 4 5 6</p>

<p>divide these 3 groups by 7… 3rd group of 6 remains because it is short 1 number
add 2 and the group will be completed</p>

<p>1 2 3 4 5 6 7
1 2 3 4 5 6 7
1 2 3 4 5 6 7
1</p>

<p>1 remainder</p>

<p>k/7 R:6
k+1/7 R:0
k+2/7 R:1</p>

<p>Just use number sense.</p>

<p>Just find a value of k for which the first is true.</p>

<p>k/7 remainder: 6
(k+2)/7 remainder: 8
remainder must be < 7
take that remainder and divide by 7
8/7 remainder: 1</p>

<p>That’s how I do it anyways.</p>

<p>Just do what #2 said, use 13.</p>

<p>Remainder 6.</p>

<p>6 + 2 = 8
8 - 7 = 1</p>

<p>Find a numerical value of k and go from there. Your brain is better equipped to deal with numbers, not variables. You had the right idea, you’re just not doing it fast enough. </p>

<p>The best way to do this is to usually just pick some k that is divisible by another number only once. For example, k divided by 7 = something with remainder 6. Let’s assume 7 goes into k only once:</p>

<p>k = 1 x 7 + 6 = 13</p>

<p>Quickly verify this is true, 13/7 = 1 r 6. </p>

<p>Now we can proceed with 13 as our value for k. k + 2 = 15. What is the remainder when 15 is divided by 7? Now that we’re dealing with numbers, this is a cinch.</p>

<p>15/7 = 2 r 1</p>

<p>Remainder = 1</p>

<p>This is the easiest way:

  1. Find a variable that has reminder of 6 …
  2. add 2
    3.divide and find reminder.</p>

<h1>colllege kid 92: I think you can do it by 3 ways:</h1>

<p>_ The first way (like collegeboard’s, but trickier):
k/7 and remainder is 6 -> (k+2)/7 and “remainder” is (6+2)=8 -> since the remainder when dividing 8 for 7 is 1, the answer has to be 1.</p>

<p>_ The second way (like yours):
Since they didn’t mean k had to be a fixed number, the answer is still true in a particular case. I found this style of thinking pretty helpful when doing multiple choice questions.
Quickly take a “k” which has the remainder 6 when divided by 7; say 6! (obviously!) Then you know what to do next.</p>

<p>_ The third way: This way of thinking is like when you fill in each cell of a table with many rows and only 7 columns. You have to fill out each row, one after another. Now you can see what happens when filling in 2 more cells in the row which only has one remaining.
You may call this way “a process of visualization” :cry: I found this one sometimes more effective and less time-consuming than the traditional one.</p>

<p>P.S: I don’t know why and how you got your “15” :expressionless: Or you meant k+2 (not k) could be 15?</p>

<p>This is the fastest mathematical way [modular arithmetic]:</p>

<p>k = 6 mod 7
K + 2 = 1 mod 7</p>

<p>Answer is 1. 5 seconds.</p>

<p>Answer is b - Just plug in 13 (because 13/7 has a remainder of 6). Then do 13+2/7 and you get a remainder of 1.</p>

<p>yeah i meant 13.
@jamesford, that way looks like the easiest way i could do it without goign insane! i didn’t know you could just add them to get find k. thanks1</p>