Expert: Paul Klarreich - 9/22/2008

Question
Hey Paul,
I have another question related to the previous one. (page 489 is mental note to myself where the problem is at)

If p is an integer and 3 is the remainder when 2p+7 is divided by 5, then p could be
a)2 b)3 c)4 d)5 e)6

Using your algorithm rule :
  A=kq+r
so therefore,
2p+7=5q+r is this correct?
I don't know how to eliminate q to find the possible value of p. Please Help. Thanks in advance.

Answer
Questioner:   Emily
Category:  Advanced Math
Private:  No
 
Subject:  Another find p question
Question:  Hey Paul,
I have another question related to the previous one. (page 489 is mental note to myself where the problem is at)

If p is an integer and 3 is the remainder when 2p+7 is divided by 5, then p could be
a)2 b)3 c)4 d)5 e)6

Using your algorithm rule :
 A=kq+r
so therefore,
2p+7=5q+r is this correct?
I don't know how to eliminate q to find the possible value of p. Please Help. Thanks in advance.
..................................................
Hi, Emily,

Since you are doing integer arithmetic, you don't just 'solve' or eliminate, because you have to get whole number answers.  Just test the cases:


p = 2:  2p + 7 = 2(2) + 7 = 11 == 1 remainder

p = 3:  2p + 7 = 2(3) + 7 = 13 == 3  << looks like the solution

p = 4:  2p + 7 = 2(4) + 7 = 15 == 0

p = 5:  2p + 7 = 2(5) + 7 = 17 == 2

p = 6:  2p + 7 = 2(6) + 7 = 19 == 4


[it ends there, because if you try any (p + 5), you get:  2p + 10 + 7, which always gives the same remainder on division by 5 as does  2p + 7. [Do you understand why?]