<p>hey guys, so I am really confused about this math question and collegeboard didn't really do a good job of explaining it. Can someone help me? </p>
<ol>
<li>When the positive integer K is divided by 7, the remainder is 6. What is the remainder when k+2 is divided by 7?</li>
</ol>
<p>a) 0
b) 1
c) 2
d) 3
e) 4</p>
<p>thanks</p>
<p>plug in a random number for K.
Let K = 13. (it satisfies the condition neatly)</p>
<p>K+2 -> 13+2 = 15
15/7 , remainder is 1
so B) is correct</p>
<p>thank you very much :)</p>
<p>Never forget the power of plug-in. :)</p>
<p>An alternate explanation would be that since when k is divided by 7 and the remainder is 6, the algebraic equation is k=7X+6. Now, since the question is asking for k+2, add 2 to the other side of the equation as follows –>k+2=7X +6+2–>7X +8. Thus, you now have,</p>
<p>–> k+2=7X+8. Now, factor the right handed side of the equation as follows.
–>k+2=7(X+1)+1. Since 7(X+1) is any multiple of 7, 1 is the remainder.</p>
<p>You could also do this neatly with modular arithmetic, which is essentially what nocensure did:</p>
<p>k = 6 (mod 7)
k + 2 = 8 (mod 7)
k + 2 = 1 + 7 = 1 + 0 = 1 (mod 7)</p>
<p>which means that the remainder when k + 2 is divided by 7 is 1.</p>
<p>noncensure, yeah but who would use that when a 2 second plug would do the job ? lol. but knowing how to problems in multiple ways is certainly helpful. not saying that setting up the algebraic eqn is bad or anything, it’s definitely good practice.</p>
<p>True, a plug in would be much more efficient, but as square said, it would be useful to know the theory for a potentially harder question in which plugging in, in contrast, would just take too long.</p>
<p>take a look at this question, how can i approach it
let the function p be defined by p(x)=a(x-k)^2, where a and k are positive constants for what value of x will the function p have its minimum value
-a
-k
0
a
k</p>
<p>the answer is k</p>
<p>take a look at this question, how can i approach it
let the function p be defined by p(x)=a(x-k)^2, where a and k are positive constants for what value of x will the function p have its minimum value
-a
-k
0
a
k</p>
<p>the answer is k </p>
<p>Plug in numbers (you don’t need to, but it helps me)</p>
<p>You need to find what you can plug in for x to give p(x) its lowest value. so</p>
<p>a=1
k=3</p>
<p>A. p(x)=1(-1-3)^2…1(-4)^2…16
B. p(x)=1(-2-3)^2…1(-5)^2…25
C. p(x)=1(0-3)^2…1(-3)^2…9
D. p(x)=1(1-3)^2…1(-2)^2…4
E. p(x)=1(2-3)^2…1(0)^2…1 so E is your answer</p>
<p>It doesn’t matter what numbers you plug in, so i chose a=1 to avoid having to multiply it, and chose 3 randomly.</p>
<p>As I was talking about, this is the case where plugging in would actuall take up more time. Think about it this way, the smallest value of any number that is squared is 0, if you have -4 for example, the answer is still 16 and even if you have 4, the answer is sixteen. 0^2 however gives you 0.</p>
<p>Thus for what value of x in (x-k)^2 will give you 0? The answer is thus
k since (k-k)^2 will give you zero. Since any a*0 is 0, the answer is k.</p>
<p>or take the derivative p’ = 2a(x-k)= 0 , x=k. lol jk</p>
<p>key thing here is to see that since both a and k are positive constants(they cannot be changed). Now ask yourself: Can this function p(x) have negative values? No, because it’s ( )^2 and a is positive. If negative values aren’t a choice, check if p(x) can be 0, since 0 is next smaller number in line.</p>