<p>It is from the blue book but I still don’t understand so I am looking for some extra help.</p>
<li>If n and p are integers greater than 1 and if p is a factor of both n + 3 and n + 10, what is the value of p?
(A) 3
(B) 7
(C) 10
(D) 13
(E) 30
The answer is B.</li>
</ol>
<p>What if i plug in 2 for n that would give me 5 and 12 as the numbers… 7 is definitly not a factor of 5 and 12???</p>
<p>Hmm, I agree something if fishy if you typed it out correctly.</p>
<p>A factor has to be smaller than the digit for it to be even possible… so anything greater than 5 is not possibly an answer.</p>
<p>yea i typed it out 100% correct… im looking at the responses from past posters and the answers are like take n+10 and minus n+3 and you get n+7… well why do you need to do this… this is a classic plug in question and what if n is 2 that still makes the thing 9…</p>
<p>awww!</p>
<p>The question probably meant to ask, what is a POSSIBLE value of P.</p>
<p>A possible value would be 7, when n = 4.</p>
<p>However, since you said the answer is D, it’s not 13.</p>
<p>oh sorry i posted the answer wrong… it is supposed to be b i will fix that… but the wording is 100% correct</p>
<p>Well basically, rather than interpreting n as a fixed value, I think their suggesting that n can be ANY value. Bad wording though.</p>
<p>This problem is not worded incorrectly, but it is nasty…</p>
<p>You could play around until you find numbers that work. But it can’t be any old numbers – they have to fit the problem. The first set that I found were:</p>
<p>Use n=4. Then n+3=7 and n+10 = 14, both of which are divisible by 7.</p>
<p>As you have pointed out, n=2 did not work…5 and 12 have no common factors.</p>
<p>Other numbers that work: n=11 which makes n+3=14 and n+10=21, again both divisible by 7.</p>
<p>There is a “math” reason this works:</p>
<p>The key is to notice that n+10 is 7 more than n+3.</p>
<p>If p is a factor of n+3, it goes into n+3 an integer number of times.</p>
<p>If it also is a factor of n+10, it also goes into n+10 an integer number of times, in fact one more time than it goes into n+3. That difference is 7, so 7 must be the number you are dividing by…</p>
<p>Even as I read what I just wrote, it barely makes sense to me. Yet another occasion where it’s just better to make up numbers.</p>
<p>pckeller provided a good math reason, though i think i would have said it differently. p goes into n+3 and n+10, therefore it goes into their difference which is (n+10)-(n+3)=7. why? because by being a factor it goes into each a certain integer number of times (e.g. 1,2,3…) so you just take the difference between how many it goes into one and the other. that gives you have how many it goes into their difference.</p>
<p>for example, if 5 goes into 30 exactly 6 times and into 40 exactly 8 times then you know right off the bat that it goes into (40-30) exactly 8-6=2 times. look at it this way: 2 is the additional bit you needed to get to 40 from 30. </p>
<p>since p goes into 7, by definition it is a factor of 7. the only factors of 7 are 1 and 7, and p≠1 by assumption in the question, therefore p=7. there is actually only one p value that works which is why the question can say “the value of p.”</p>
<p>regarding wording, it is perfectly fine you just have to satisfy all the if’s first. your example of n=2 does not satisfy the if’s.</p>
<p>i do understand how it works (at the time of my practice test i didn’t), but if they state that n is an integer greater than 1, technically that means n is any integer greater than one… they never set out other restrictions such as n must equal 4, 11, 18 (this would give the question away). Yes as you guys showed there is a way to find the answer but by stating it the way they did it does technically mean any number above 1 p is a multiple of what?</p>
<p>This is a similar idea to that of pckeller.
If p is a factor of n+10 then (n+10)/p must give an integer answer</p>
<p>(n+10)/p = (n+3+7)/p = [(n+3)+7]/p = [(n+3)/p] + [7/p]</p>
<p>Since p is already a factor of n+3, (n+3)/p must be an integer So, for the final value of (n+10)/p to be an integer ,p/7 must also be an integer and thus p will also be a factor of 7.
Hence p=1,7
But since p>1, p must equal 7.</p>
<p>P= 7 ,N = 11 because 7 is factor of 11+3=14 and 11+10 =21 .THis was easy</p>
<p>
you cannot just single out one condition. one condition may seem to be talking about p but it really involves both n and p. “If n and p are integers greater than 1 and if p is a factor of both n + 3 and n + 10,” could be worded differently: “if n>1 and p>1, and if n+3 and n+10 are integer multiples of p,…” maybe that strikes you more as restricting your n values for given p values? you can’t start a problem that has premises if the premises are not satisfied, no matter how that is done.</p>
<p>…but if they state that n is an integer greater than 1, technically that means n is any integer greater than one…</p>
<p>@Ifii2, you are not correct in your assumption. It does not mean all numbers greater than 1 are the set of numbers equal to n. n is actually in the example, a specific number or set of numbers, just one that you are not certain of the value, and not all numbers. I do not think there is any way to do the problem other than as stated in post 8 or 10 above. You can never really pin down what n is, since as others have stated, n can be 4, 11, 18 etc. a repeating pattern of 4 plus multiples of 7.</p>