19. If M and N are positive integers that have remainders of 1 and 3, respectively, when divided by 6, which of the following could NOT be a possible value of M+N?
(A) 86
(B) 52
(C) 34
(D) 28
(E) 10
answer is A
Would someone please be able to help me with this one?
PS Remainder question
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The samllest possible values for M and N are 1 and 3.
1+3=4 (irrelevant)
then you have 7 and 9 (just add 6 to 1 and 3)
7+3 = 10, so E is out
9+1 also =10
7+9=16 (irrelevant)
next you have 13 and 15
13+15 = 28, so D is out, no bother to try other combinations as they will give a value smaller than 28 ant the remaining answers are higher.
next is 19 and 21
19+21=40
19+15=34(also =13+21) C is out
next is 25 and 27
25+27 = 52 so B is out, only A remains
A is the answer
1+3=4 (irrelevant)
then you have 7 and 9 (just add 6 to 1 and 3)
7+3 = 10, so E is out
9+1 also =10
7+9=16 (irrelevant)
next you have 13 and 15
13+15 = 28, so D is out, no bother to try other combinations as they will give a value smaller than 28 ant the remaining answers are higher.
next is 19 and 21
19+21=40
19+15=34(also =13+21) C is out
next is 25 and 27
25+27 = 52 so B is out, only A remains
A is the answer
- givemeanid
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M = 6x + 1
N = 6y + 3
So, M+N = 6(x+y) + 4
M+N-4 = 6(x+y) = a multiple of 6.
Scan the answer choices and subtract 4 from each of them. The resulting number should not be a multiple of 6.
1. 86 - 4 = 82. Bingo!
Try out other choices to see that when 4 is subtracted, they result in a multiple of 6.
N = 6y + 3
So, M+N = 6(x+y) + 4
M+N-4 = 6(x+y) = a multiple of 6.
Scan the answer choices and subtract 4 from each of them. The resulting number should not be a multiple of 6.
1. 86 - 4 = 82. Bingo!
Try out other choices to see that when 4 is subtracted, they result in a multiple of 6.
So It Goes
- givemeanid
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Always always remember the 2-min rule. MGMAT CATs have the rule tattooed in my brain. If you think the question is in your alley, then you SHOULD be able to crack it in and around 2-min. If not, then either the question is out of your ballpark (in which case, you make an educated guess and move on) or you are taking a longer route and their exists a quicker way of solving the problem.
I think Stacey's posts keep emphasizing the part that you should re-solve the problems that you do during practice even if you got them right to make sure you are doing the problem the fastest way possible.
Remember the problem in (I believe) OG that gives you a co-ordinates of a point in 2nd quadrant on the semi circle with center on the origin and asks you to find one of the co-ordinates of a point in first quadrant on the circle? When I came across that problem the first time, I did solve it and moved on. However, later on, after using the above mentioned advice, I redid the problem and figured out at least 2 other ways which were quicker than my original way of doing the problem. Lesson learned and locked up in the vault.
I think Stacey's posts keep emphasizing the part that you should re-solve the problems that you do during practice even if you got them right to make sure you are doing the problem the fastest way possible.
Remember the problem in (I believe) OG that gives you a co-ordinates of a point in 2nd quadrant on the semi circle with center on the origin and asks you to find one of the co-ordinates of a point in first quadrant on the circle? When I came across that problem the first time, I did solve it and moved on. However, later on, after using the above mentioned advice, I redid the problem and figured out at least 2 other ways which were quicker than my original way of doing the problem. Lesson learned and locked up in the vault.
So It Goes