<p>If S is the sum of x consecutive integers, then S must be even if x is a multiple of</p>
<p>(A) 6
(B) 5
(C) 4
(D) 3
(E) 2</p>
<p>Answer is C, but how do you exactly do this the right way?</p>
<p>ok lets try all the choices</p>
<p>if A 6 consecutive integers</p>
<p>2+3+4+5+6+7 = odd</p>
<p>if B
(it could be even but not always)
3+4+5+6+7 = 25 odd</p>
<p>if D </p>
<p>2+3+4 =9 odd</p>
<p>2 integers</p>
<p>2+3 = 5 odd</p>
<p>now we've found an exception to each of these choices... But, now to show you why 4 always works!</p>
<p>now since the integers are consecutive.... we will have 2 even integers and 2 odd integers.</p>
<p>odd +odd = even integer
&
even + even = even integer</p>
<p>and even + even = even :)</p>
<p>So for 4 no matter the starting integer it will always be even because of the even odd property</p>
<p>since it's x consecutive integers, then x could be 3 in that it's 1+2+3. And that's even, but it's not multiple of 4h</p>
<p>hmmm i just got confused on the question.... i have no clue whats its asking now...</p>
<p>need back up</p>
<p>Let n be the first integer. The subsequent integers are (n+1), (n+2), (n+3), (n+4), (n+5), and so on.</p>
<p>If x=6, then S=n+(n+1) +(n+2)+(n+3)+(n+4)+(n+5)=6n+15, which is not always even (for example, when n=1, S=6n+15=6(1)+15=21).</p>
<p>If x=5, then S=n+(n+1) +(n+2)+(n+3)+(n+4) =5n+10, which is not always even (for example, when n=1, S=5n+15=5(1)+10=15).</p>
<p>If x=4, then S=4n+6, which is always even no matter what n is. Furthermore, a multiple of an even number is always an even number.</p>
<p>If x=3, then S=3n+3, which is not always even.</p>
<p>If x=2, then S=2n+1, which is not always even.</p>
<p>So the only correct answer is x=4.</p>
<p>And by the way, the keywords in the question are "... must be...".</p>
<p>EDITED------------------</p>
<p>Good job goldtx. Excellent way of working it out. Surprised i didnt think about how to write all the possibilities in equation form. Trial and error would take up so much time.</p>
<p>The sum of two consecutive integers must be odd since even + odd = odd. Because S has to be even, we know that we need odd + odd = even, so we need 4 consecutive integers. Therefore x could be 4 or 8 which is just[(even+odd)4 = even] or any other multiple of 4. </p>
<p>We're dealing with x integers, no need to get into x + (x+1) + (x+2)... unless you want to investigate using numbers (it'll only get more confusing).</p>
<p>Or what backfire mentioned at the end.</p>
<p>Before answering a question, it is always helpful to know how much time you have to answer it. This type of question should be at the end of the section, and one could assume to have about one minute to solve it. </p>
<p>The first step is to pay close attention to the reasoning aspect. What is asked of us? Questions that SEEM confusing typically have a pretty simple answer, albeit a lengthier process. However, do not let the problem scare you! Take 15 seconds to 'really" get the problem and use an example to move from the abstract to the concrete.</p>
<p>Students who are good at writing equations should follow their instincts. However, on the SAT, questions such as this one, or probability problems can be solved by writing them out. </p>
<p>In this case, my personal inclination would have been to write it out, as I can use VERY easy numbers. </p>
<p>This is how my sheet would look like:</p>
<p>1,2,3,4,5,6 ----21 no
1,2,3,4,5, ----15 no
1,2,3,4, ----10 yes 5,6,7,8 ---- 26 --36 Yes
1,2,3, ----6 4,5,6 ---- 15 No
1,2, -3 No</p>
<p>The process is very fast and very simple. Three answers were eliminated immediately (pretty typical for the SAT) and the second elimination round did not take more than a few seconds.</p>
<p>If S is the sum of x consecutive integers, then S must be even if x is a multiple of</p>
<p>(A) 6
(B) 5
(C) 4
(D) 3
(E) 2</p>
<p>The answer is C because for every 4 consecutive integers, you will have 2 even integers and 2 odd integers, therefore 2 will always be a factor of their sum. Simple.</p>
<p>i know...I thought x could be any random amount of integers as long as it's multiple of 4. Darn</p>
<p>Not to get esoteric, the sum of any two odd numbers is even and the sum of any two even numbers is even,therefore, if you want to add x consecutive numbers to be even two of them must be odd. This lends itself to four consecutive numbers.
ex.
1234; 2345,3456: all even which is true for any four consecutive numbers.
Also note for fun, 1+3=4; 3+5=8; 5+7=12; 7+9=16</p>
<p>xiggi -- I agree, simple plugging in (as unscientific as it sounds) is often the fastest way on the SAT.
I would shave off .5 seconds going from the bottom up:
(E) - no,
(D) - no,
(C) - yes.
Can't we stop with (C), since only one answer can be correct?</p>
<p>Edit.
Another .4 seconds off: we could use 0 as the first number.
(E): 0+1 = 1 -- no
(D): 0+1+2 = 3 -- no
(C): 0+1+2+3 = 6, 0+1+2+3+4 = 10. YES!</p>