<li>If p is an integer and 3 is the remainder when 2p+7 is divided by 5, then p could be</li>
</ol>
<p>a 2
b 3
c 4
d 5
e 6</p>
<p>I know you can “plug-in the answer,” but is there a faster, perhaps formulaic, way?</p>
<p>Yes there is. set it up as an equation whereby 2P+7/(5) equals 3.</p>
<p>To be divisible by 5 and leave a remainder of 3, a number has to end in 3 or 8.</p>
<p>This leaves only 13 and 18 as choices for 2p + 7 or 2 p being either 11 or 6. Obviously, only 6 would work, so p=3.</p>
<p>When dealing with such small integers, the best option is really to use plug-in. It really takes no time to try a couple of them. Setting up an equation is a waste of time and adds the risk of making a mistake.</p>
<p>
</p>
<p>Slight problem. That equation would be solving for when 2p + 7 = 15, not when there is a remainder of 3.</p>
<p>"Remainder" questions are fairly common on SAT.</p>
<p>Modular arithmetics could be very useful, even in super simple questions like this one.</p>
<p>"The Math Forum" and "Cut-the-knot" are on my top ten math sites list.
They give a very good intro:</p>
<h1><a href="http://www.cut-the-knot.org/blue/Modulo.shtml%5B/url%5D">http://www.cut-the-knot.org/blue/Modulo.shtml</a></h1>
<p>2p + 7 = 3 (mod 5)
2p = -4 (mod 5)
p = -2 (mod 5)
p could be
5 - 2 = 3 - bingo! we can stop now.<br>
Let's just find a few more:
10 - 2 = 8
or
15 - 2 = 13
or
5k - 2, in general.</p>
<p>++++++++++++++++++++++++++
If you don't feel like learning the whole new math section just for one type of questions, there are a couple of useful remainder facts you should fish for around in your books.</p>
<h1>They are still from modulo arithmetics, only without the use of formulas.</h1>
<p>After some practice you'd do this kind of things in your head in seconds.</p>
<p>Let's just make a question a tiny bit more difficult by shifting the answers 2 up:
a 4
b 5
c 6
d 7
e 8</p>
<p>Important thing not to be missed:
answer choices are CONSEQUITIVE numbers.
That tells us that
a remainder from dividing (2p+7) by 5 will increase by 2 after going one answer choice up.</p>
<p>[ This part should be done automatically without thinking, just based on the fact 2(p+1)=2p+2 - you should not even write it.
OK, here's the formal proof:
If 2p+7 = 5k+r, where r is a remainder from dividing (2p+7) by 5, then
2(p+1) + 7 = 2p+7 + 2 = 5k + r+2. ]</p>
<p>Plug in answer (a):
2*4 + 7 = 15, remainder 0 when divided by 5.</p>
<p>For the next 4 answer choices remainders will be
0+2=2
2+2=4
4+2=6=1 (don't ask why; think)
1+2=3 - this is our guy. Answer (e).</p>
<p>Plug-in method works beautiful when combined with time saving techniques.</p>
<p>You just confused the hell out of me, gcf101.</p>
<p>Sorry, Flipsta! Did not mean to.</p>
<p>I guess, I overdone a tad with the brevity.</p>
<p>21 = 5<em>4 + 1, so 1 is a remainder from dividing 21 by 5.
Since 21 = 5</em>3 + 6, we can say that 6 is also a remainder from dividing 21 by 5, just not the smallest possible.
1 = 6 - 5.
We can even say now that remainders 1 and 6 are equivalent in a sense.</p>
<p>We can say anything we want!!
It's 3:45 a.m. for crying out loud, and I am cooked!</p>
<p>If r is a remainder from dividing (2p+7) by 5, then
for different values of p, corresponding remainders r will be:</p>
<p>p=4, 2p+7=15, r=0</p>
<p>p=5, 2p+7=17, r=2</p>
<p>p=6, 2p+7=19, r=4</p>
<p>p=7, 2p+7=21, r=1 (or r=6)</p>
<p>p=8, 2p+7=23, r=3.</p>
<p>Each following remainder is 2 more than preceding one;
r=6 is replaced with the equivalent r=1.</p>
<p>In other words, if p goes through the values of
4 - 5 - 6 - 7 - 8,
remainder r takes the values of
0 - 2 - 4 - 1 - 3.</p>
<p>I am afraid I am more confused now then you are.
Or you are still ahead?
Roger and out.</p>
<p>if (2p+7)/5 has a remainder of 3, then [(2p+ 7)-3]/5 => (2p+4)/5 => 2(p+2)/5 => (p+2)/5 has a remainder of ZERO. So (p+2) ends either 0 or 5. you do not have to plug in every answer to save time.</p>
<p>right, same as above (post #5)
p = -2 (mod 5) <=></p>
<p>p+2 = 0 (mod 5)</p>