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
OA is A. I had E. Please explain and any tricks to solve these quickly would be appreciated!
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Mclaughlin
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parallel_chase
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M= 6x+1
N= 6x+3
M+N=12x+4
Therefore any value which is divided by 12 does not give a remainder 4 is the answer.
86 = 12*7+2
52 = 12*4+4
34 = 12*6.8+4
28 = 12*2+4
10 = 12*0.8+4
Hence A.
N= 6x+3
M+N=12x+4
Therefore any value which is divided by 12 does not give a remainder 4 is the answer.
86 = 12*7+2
52 = 12*4+4
34 = 12*6.8+4
28 = 12*2+4
10 = 12*0.8+4
Hence A.
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pepeprepa
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Chase I think what you wrote is dangerous
"M= 6x+1
N= 6x+3
M+N=12x+4 "
It is possible the quotient is the same for M and N but we are not sure of that.
You can write
M+N=6(x+y)+4
And you check as Chase checked in his explanation.
"M= 6x+1
N= 6x+3
M+N=12x+4 "
It is possible the quotient is the same for M and N but we are not sure of that.
You can write
M+N=6(x+y)+4
And you check as Chase checked in his explanation.
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parallel_chase
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You are absolutely right "pepeprepa", I knew something was wrong. Thanks for spotting the error.pepeprepa wrote:Chase I think what you wrote is dangerous
"M= 6x+1
N= 6x+3
M+N=12x+4 "
It is possible the quotient is the same for M and N but we are not sure of that.
You can write
M+N=6(x+y)+4
And you check as Chase checked in his explanation.
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sudhir3127
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good spot pepeprepa.. i used almost the same approach ... but it was faster for me .. no equations though..
We know that any number when divided by 6 shud have a remiander 4.
i just subtracted 4 from each number and except A ... every other option is divisble by 6. hence no remainder...
hope its a right approach....
We know that any number when divided by 6 shud have a remiander 4.
i just subtracted 4 from each number and except A ... every other option is divisble by 6. hence no remainder...
hope its a right approach....
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beeparoo
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Likewise, I also calculated that M + N = 6(x + y) + 4.Mclaughlin wrote: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
OA is A. I had E. Please explain and any tricks to solve these quickly would be appreciated!
Thus, sum of M + N is 4 more than a multiple of 6.
I agree that B, C, and D are easily eliminated by observation.
While I can clearly see that A is NOT a possible value for M + N, I still question how E (10) is a possible value for M + N.
M + N = 10 = 6(x + y) + 4
x + y = 1
But since x and y are each quotients of positive integers, should they not also be positive integers as well? They cannot be partial numbers (like 2/3 and 1/3), nor can they be negative numbers (like, 2 and -1).
How can I confidently determine that M + N may be = 1? Right now, it seems NOT possible.
GMAT obsession begone - girl needs her social life back.
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parallel_chase
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Here is a definition to help you understand remainders better.
"when the result of the division of two integers cannot be expressed with an integer quotient, the remainder is the amount "left over."
The quotient can have decimal values.
if you divide 1 by 3, the quotient will be .333.... and this will always leave a remainder of 1.
Similarly if you divide 10 by 6, the quotient can be 1.6666 and this will always leave a remainder of 4.
Hope this helps.
"when the result of the division of two integers cannot be expressed with an integer quotient, the remainder is the amount "left over."
The quotient can have decimal values.
if you divide 1 by 3, the quotient will be .333.... and this will always leave a remainder of 1.
Similarly if you divide 10 by 6, the quotient can be 1.6666 and this will always leave a remainder of 4.
Hope this helps.
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beeparoo
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Oh yeesss... I feel like I've just had an "of course!" moment.parallel_chase wrote:Here is a definition to help you understand remainders better.
"when the result of the division of two integers cannot be expressed with an integer quotient, the remainder is the amount "left over."
The quotient can have decimal values.
if you divide 1 by 3, the quotient will be .333.... and this will always leave a remainder of 1.
Similarly if you divide 10 by 6, the quotient can be 1.6666 and this will always leave a remainder of 4.
Hope this helps.
Thanks a lot!
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Ian Stewart
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When we talk about quotients in the context of quotients and remainders, we almost always mean the 'integral quotient'. An integral quotient is never a decimal, always an integer. When you divide n by d, you can always find (assuming n and d are positive integers), unique integers q and r so that:parallel_chase wrote:Here is a definition to help you understand remainders better.
"when the result of the division of two integers cannot be expressed with an integer quotient, the remainder is the amount "left over."
The quotient can have decimal values.
if you divide 1 by 3, the quotient will be .333.... and this will always leave a remainder of 1.
Similarly if you divide 10 by 6, the quotient can be 1.6666 and this will always leave a remainder of 4.
Hope this helps.
n = qd + r, where 0 <= r < d
q is called the integral quotient, or more usually, just the 'quotient';
r is called the 'remainder'.
When you divide 10 by 6, the quotient is 1 and the remainder is 4, because
10 = 1*6 + 4
When you divide 1 by 3, the quotient is 0 and the remainder is 1, because
1 = 0*3 + 1
To answer beeparoo's question, M+N can equal ten because quotients can be equal to zero (in your notation, one of x or y could be zero, and the other could be one).
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
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