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
is there any other way to solve tis problem??
what would be the fastest way to solve such problems..??
OA=A[spoiler][/spoiler]
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any other method
This topic has expert replies
hi,
first,are you sure the answer is A.Because it seems to me like the right answer should be E.let me explain.
when M is divided by 6,remainder is 1,so let's have an equation
M/6=x + 1
When N is divided by 6,remainder is 3,so
N/6=y + 3
When we cross-multiply,we get M=6x +6 and N=6y+18
Therefore M+N=9x+6y+24.
Looking at the answer choices,we see that the most unlikely choice is 10 coz the above equation already tells us that M+N must be greater than or equal to 24.Hope this helps.
first,are you sure the answer is A.Because it seems to me like the right answer should be E.let me explain.
when M is divided by 6,remainder is 1,so let's have an equation
M/6=x + 1
When N is divided by 6,remainder is 3,so
N/6=y + 3
When we cross-multiply,we get M=6x +6 and N=6y+18
Therefore M+N=9x+6y+24.
Looking at the answer choices,we see that the most unlikely choice is 10 coz the above equation already tells us that M+N must be greater than or equal to 24.Hope this helps.
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niraj_a
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i believe it is A and here's my thought process,
if remainder of M is 1, then M could be 6, and similarly N could be 9. then if 9+7 = 16 is divided 6, then remainder we get is 4.
so if we backsolve with the answer choices, any answer choice that doesn't yield a remainder 4 when divided by 6 should be our answer.
so 86/6 meets that logic so A.
if remainder of M is 1, then M could be 6, and similarly N could be 9. then if 9+7 = 16 is divided 6, then remainder we get is 4.
so if we backsolve with the answer choices, any answer choice that doesn't yield a remainder 4 when divided by 6 should be our answer.
so 86/6 meets that logic so A.
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Ian Stewart
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This starts out well, but there is no reason why x+y must be larger than 2. There is nothing wrong with having a quotient of 0 in division: when you divide 5 by 6, for example, the quotient is 0 and the remainder is 5. x and y are the quotients here, and they can certainly be equal to 0.duongthang wrote:M=6x+1
N=6y+3
M+N=6(x+y)+ 4
x+y must be bigger than 2, M+N must bigger than 16
E is the right answer
That said, your approach is good to begin with. We know
M+N = 6(x+y) + 4
That means that M+N must be 4 larger than a multiple of 6, or in other words, that the remainder must be 4 when you divide M+N by 6. This is essentially what niraj found above by picking numbers. A is the only answer that gives a remainder different from 4 when you divide by 6.
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tokaitalbo
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For me, doing this out seems the easiest, and fastest.
if the remainder is 1, then M can only be 1,7,13,19,25
if the remainder is 3, then N can only be 3,9,15,21,27
start at the bottom (smallest) and use POE
cannot be E: 1+9=10
cannot be D: 27+1=28
cannot be C: 27+7=34
cannot be B: 25+27=52.
Answer must be A.
if the remainder is 1, then M can only be 1,7,13,19,25
if the remainder is 3, then N can only be 3,9,15,21,27
start at the bottom (smallest) and use POE
cannot be E: 1+9=10
cannot be D: 27+1=28
cannot be C: 27+7=34
cannot be B: 25+27=52.
Answer must be A.
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yvichman
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Can't you just divide each of the answer choices given, by 6 to find out if they give a remainder of 4 (3+1 as stated in the question). The one that does not have a remainder of 4 would give you the correct answer.
Is this an approach that would not work all the time?
Is this an approach that would not work all the time?
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Ian Stewart
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Yes, that's perfect. My post above is identical in method- I just wanted to demonstrate why the remainder needs to be equal to 4.yvichman wrote:Can't you just divide each of the answer choices given, by 6 to find out if they give a remainder of 4 (3+1 as stated in the question). The one that does not have a remainder of 4 would give you the correct answer.
Is this an approach that would not work all the time?
